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Calculate Derivatives – Problems & Solutions

Are you working to calculate derivatives in Calculus? Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself.

Jump down this page to: [Power rule: $x^n$] [Exponential: $e^x$] [Trig derivs] [Product rule] [Quotient rule] [More problems & University exam problems]
[Chain rule (will take you to a new page)]

Update: We now have a much more step-by-step approach to helping you learn how to compute even the most difficult derivatives routinely, inclduing making heavy use of interactive Desmos graphing calculators so you can really learn what’s going on. Please visit our Calculating Derivatives Chapter to really get this material down for yourself.

It’s all free, and waiting for you! (Why? Just because we’re educators who believe you deserve the chance to develop a better understanding of Calculus for yourself, and so we’re aiming to provide that. We hope you’ll take advantage!)

If you just need practice calculating derivative problems for now, previous students have found what’s below super-helpful. And if you have questions, please ask on our Forum! It’s also free for your use.

Matheno Essentials: Derivatives and Rules Summary
You can always access our Handy Table of Derivatives and Differentiation Rules via the Key Formulas menu item at the top of every page.
CLICK TO VIEW SUMMARY

Power of x

$$\frac{d}{dx}\text{(constant)} = 0 \quad \frac{d}{dx} \left(x\right) = 1 $$ $$\frac{d}{dx} \left(x^n\right) = nx^{n-1} $$

Exponential

\begin{align*} \frac{d}{dx}\left( e^x \right) &= e^x &&& \frac{d}{dx}\left( a^x \right) &= a^x \ln a \\ \\ \end{align*}

Trigonometric

\begin{align*} \frac{d}{dx}\left(\sin x\right) &= \cos x &&& \frac{d}{dx}\left(\csc x\right) &= -\csc x \cot x \\ \\ \dfrac{d}{dx}\left(\cos x\right) &= -\sin x &&& \frac{d}{dx}\left(\sec x\right) &= \sec x \tan x \\ \\ \dfrac{d}{dx}\left(\tan x\right) &= \sec^2 x &&& \frac{d}{dx}\left(\cot x\right) &= -\csc^2 x \end{align*}

Notice that a negative sign appears in the derivatives of the co-functions: cosine, cosecant, and cotangent.

Constant Factor Rule

Constants come out in front of the derivative, unaffected: $$\dfrac{d}{dx}\left[c f(x) \right] = c \dfrac{d}{dx}f(x) $$

For example, $\dfrac{d}{dx}\left(4x^3\right) = 4 \dfrac{d}{dx}\left(x^3 \right) =\, … $


Sum of Functions Rule

The derivative of a sum is the sum of the derivatives: $$\dfrac{d}{dx} \left[f(x) + g(x) \right] = \dfrac{d}{dx}f(x) + \dfrac{d}{dx}g(x) $$

For example, $\dfrac{d}{dx}\left(x^2 + \cos x \right) = \dfrac{d}{dx}\left( x^2\right) + \dfrac{d}{dx}(\cos x) = \, …$


Product Rule

\begin{align*} \dfrac{d}{dx}(fg)&= \left(\dfrac{d}{dx}f \right)g + f\left(\dfrac{d}{dx}g \right)\\[8px] &= {\small\Big[\text{ (derivative of the first) } \times \text{ (the second) }\Big] + \Big[\text{ (the first) } \times \text{ (derivative of the second)}\Big]} \end{align*}

IV. Quotient Rule

\begin{align*} \dfrac{d}{dx}\left(\dfrac{f}{g} \right) &= \dfrac{\left(\dfrac{d}{dx}f \right)g – f\left(\dfrac{d}{dx}g \right)}{g^2} \\[8px] &={\small\dfrac{{\Big[\text{(derivative of the numerator) } \times \text{ (the denominator)}\Big] – \Big[\text{ (the numerator) } \times \text{ (derivative of the denominator)}}\Big]}{\text{all divided by [the denominator, squared]}}} \end{align*}

Many students remember the quotient rule by thinking of the numerator as “hi,” the demoninator as “lo,” the derivative as “d,” and then singing

“lo d-hi minus hi d-lo over lo-lo”

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I. Power Rule
Most frequently, you will use the Power Rule: \[\bbox[yellow,5px]{\dfrac{d}{dx}\left(x^n\right) = nx^{n-1}}\] This is just a fancy, compact way of capturing \begin{align*} \dfrac{d}{dx}(x) &= 1x^0 = 1 \\ \dfrac{d}{dx}\left(x^2\right) &= 2x\\ \dfrac{d}{dx}\left(x^3 \right) &= 3x^2\\ \dfrac{d}{dx}\left( x^4 \right)&= 4x^3 \\ \vdots\\ \dfrac{d}{dx}\left(x^{157}\right) &= 157x^{156}\\ & \text{etc.}& \end{align*} The rule works just the same for negative exponents: \begin{align*} \dfrac{d}{dx}(x^{-1}) &= -1x^{-2} \\ \dfrac{d}{dx}\left(x^{-2}\right) &= -2x^{-3}\\ \dfrac{d}{dx}\left(x^{-3} \right) &= -3x^{-4}\\ & \text{etc.}& \end{align*} The rule also captures the fact that the derivative of a constant ($c$) is zero: $${\bf\text{Constant, c:}} \qquad \dfrac{d}{dx}(c) = \dfrac{d}{dx}cx^0 = 0$$ Finally, because $\sqrt{x}$ comes up so frequently, even though it's easy to compute (as we will below), it's worth memorizing $${\bf\text{Square root:}} \qquad \dfrac{d}{dx}\sqrt{x} = \dfrac{1}{2\sqrt{x}}$$
Try your hand at using the Power Rule in the following problems. As always, the complete solution is immediately available by clicking "Show/Hide Solution."
Power Rule Differentiation Problem #1
Differentiate $f(x) = 2\pi$.
Show/Hide Solution
$2\pi$ is just a number: it’s a constant. And the derivative of any constant is 0: \[ \begin{align*} \dfrac{d}{dx}(2\pi) &= \dfrac{d}{dx}(\text{constant}) \\[8px] &= 0 \quad \cmark \end{align*} \]
[hide solution]
Power Rule Differentiation Problem #2
Differentiate $f(x) = \dfrac{2}{3}x^9$.
Show/Hide Solution
We’ll show more detailed steps here than normal, since this is the first time we’re using the Power Rule. Recall that $\dfrac{d}{dx}\left(x^n\right) = nx^{n-1}.$ \[ \begin{align*} \dfrac{d}{dx}\left( \frac{2}{3}x^9\right) &= \frac{2}{3} \dfrac{d}{dx}\left(x^9 \right) \\[8px] &= \frac{2}{3}\left(9 x^{9-1} \right) \\[8px] &= \frac{2}{3}(9) \left(x^8 \right) \\[8px] &= 6x^8 \quad \cmark \end{align*} \]
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Power Rule Differentiation Problem #3
Differentiate $f(x) = 2x^3 - 4x^2 + x -33$.
Show/Hide Solution
Recall that $\dfrac{d}{dx}\left(x^n\right) = nx^{n-1}.$ We simply go term by term: \[ \begin{align*} \dfrac{d}{dx}\left(2x^3 – 4x^2 + x -33 \right) &= \dfrac{d}{dx}\left( 2x^3\right) – \dfrac{d}{dx}\left( 4x^2\right) + \dfrac{d}{dx}(x) – \cancelto{0}{\dfrac{d}{dx}(33)} \\[8px] &= 2\dfrac{d}{dx}\left( x^3\right) – 4\dfrac{d}{dx}\left( x^2\right) + \dfrac{d}{dx}(x) \\[8px] &= 2 \left(3 x^2 \right) – 4 \left(2x^1 \right) + 1 \\[8px] &= 6x^2 -8x + 1 \quad \cmark \end{align*} \]
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Power Rule Differentiation Problem #4
Differentiate $f(x) = x^{1001} + 5x^3 -6x +10,687$.
Show/Hide Solution
\begin{align*} \frac{df}{dx} &= \frac{d}{dx}\left[x^{1001} + 5x^3 -6x +10,687 \right] \\[8px] &= \frac{d}{dx}\left(x^{1001}\right) + \frac{d}{dx}\left(5x^3\right) – \frac{d}{dx}(6x) + \cancelto{0}{\frac{d}{dx}(10,687)}\\[8px] &= \frac{d}{dx}\left(x^{1001}\right) + 5 \frac{d}{dx}\left(x^3\right) – 6 \frac{d}{dx}(x) \\[8px] &= 1001x^{1000} + 5(3x^2) – 6(1) \\[8px] &= 1001x^{1000} + 15x^2 – 6 \quad \cmark \end{align*}
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Power Rule Differentiation Problem #5: Sqrt(x)
Show that $\dfrac{d}{dx}\sqrt{x} = \dfrac{1}{2}\dfrac{1}{\sqrt{x}}$.
Tips iconRecall that $\sqrt{x} = x^{1/2}.$
Show/Hide Solution
Recall that $\dfrac{d}{dx}\left(x^n\right) = nx^{n-1}.$ The rule also holds for fractional powers: \[ \begin{align*} \dfrac{d}{dx}\left(\sqrt{x} \right) &= \dfrac{d}{dx}\left(x^{\frac{1}{2}} \right) \\[8px] &= \frac{1}{2}x^{\left(\frac{1}{2}\, – 1 \right)} \\[8px] &= \frac{1}{2}x^{-1/2} \\[8px] &= \frac{1}{2}\frac{1}{\sqrt{x}} \quad \cmark \end{align*} \]
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Power Rule Differentiation Problem #6
Differentiate $f(x) = \dfrac{5}{x^3}$.
Tips iconRecall that $\dfrac{1}{x^n} = x^{-n}.$
Show/Hide Solution
Recall that $\dfrac{d}{dx}\left(x^n\right) = nx^{n-1}.$ The rule also holds for negative powers: \[ \begin{align*} \dfrac{d}{dx} \left(\dfrac{5}{x^3} \right) &= 5 \dfrac{d}{dx} \left(x^{-3} \right) \\[8px] &= 5 \left((-3) x^{(-3-1)} \right) \\[8px] &= -15 x^{-4} \quad \cmark \\[8px] &= \dfrac{-15}{x^4} \quad \cmark \end{align*} \] Note that the last two lines are completely equivalent, and either would be acceptable as the answer.
[hide solution]
Power Rule Differentiation Problem #7
Differentiate $f(x) = \dfrac{1}{x^2} - \dfrac{4}{x^5}$.
Show/Hide Solution
\begin{align*} \frac{df}{dx} &= \frac{d}{dx}\left[ \dfrac{1}{x^2} – \dfrac{4}{x^5}\right]\\[8px] &= \frac{d}{dx}\left(\dfrac{1}{x^2}\right) – \frac{d}{dx}\left(\dfrac{4}{x^5}\right) \\[8px] &= \frac{d}{dx}\left(x^{-2}\right) – 4\frac{d}{dx}\left(x^{-5}\right) \\[8px] &= -2x^{-3} – 4\left(-5x^{-6} \right) \\[8px] &= -2x^{-3} +20x^{-6} = -\frac{2}{x^3} + \frac{20}{x^6} \quad \cmark \end{align*} Note: Either way of writing the last line, with negative exponents or fractions, is correct.
[hide solution]
Power Rule Differentiation Problem #8
Differentiate $f(x) = \sqrt[3]{x}\, - \dfrac{1}{\sqrt{x}}$.
Show/Hide Solution
Recall that $\dfrac{d}{dx}\left(x^n\right) = nx^{n-1}.$ \[ \begin{align*} \dfrac{d}{dx}\left(\sqrt[3]{x}\, – \dfrac{1}{\sqrt{x}} \right) &= \dfrac{d}{dx} \left(x^{1/3} \right) – \dfrac{d}{dx} \left(x^{-1/2} \right) \\[8px] &= \left(\frac{1}{3}x^{(1/3) – 1} \right) – \left(-\frac{1}{2} x^{\left(-(1/2) – 1 \right)} \right) \\[8px] &= \frac{1}{3} x^{-2/3} + \frac{1}{2}x^{-3/2} \quad \cmark \\[8px] &= \frac{1}{3}\frac{1}{\sqrt[3]{x^2}} + \frac{1}{2} \frac{1}{\sqrt{x^3}} \quad \cmark \end{align*} \] Note that the last two lines are completely equivalent, and either would be acceptable as the answer.
[hide solution]
Power Rule Differentiation Problem #9
Differentiate $f(x) = \sqrt{x}\left(x^2 - 8 + \dfrac{1}{x} \right)$.
Show/Hide Solution
We’ll learn the “Product Rule” below, which will give us another way to solve this problem. For now, to use only the Power Rule we must multiply out the terms. Recall that $x^a x^b = x^{(a+b)}.$ \[ \begin{align*} \sqrt{x}\left(x^2 – 8 + \frac{1}{x} \right) &= x^{1/2}x^2 – 8x^{1/2} + x^{1/2}x^{-1} \\[8px] &= x^{5/2}\, – 8x^{1/2}\, + x^{-1/2} \end{align*} \] We can now take the derivative: \[ \begin{align*} \dfrac{d}{dx} \left(\sqrt{x}\left(x^2 – 8 + \frac{1}{x} \right) \right) &= \dfrac{d}{dx}\left(x^{5/2}\right) – 8\dfrac{d}{dx}\left(x^{1/2} \right) + \dfrac{d}{dx} \left( x^{-1/2} \right) \\[8px] &= \frac{5}{2} x^{\left((5/2) – 1 \right)} – 8 \left(\frac{1}{2} x^{\left((1/2) – 1 \right)} \right) + \left(-\frac{1}{2} \right)x^{\left(-(1/2) -1 \right)} \\[8px] &= \frac{5}{2}x^{3/2}\, – 4 x^{-1/2} \,- \frac{1}{2} x^{-3/2} \quad \cmark \end{align*} \]
[hide solution]
Power Rule Differentiation Problem #10
Differentiate $f(x) = \left(2x^2 + 1 \right)^2$.
Show/Hide Solution
To use only the Power Rule to find this derivative, we must start by expanding the function so we can proceed term by term: \[ \begin{align*} \left(2x^2 + 1 \right)^2 &= (2x^2)^2 + 2(2x^2)(1) + 1^2 \\[8px] &= 4x^4 + 4x^2 + 1 \end{align*} \] We can now take the derivative: \[ \begin{align*} \dfrac{d}{dx}\left(2x^2 + 1 \right)^2 &= \dfrac{d}{dx}\left(4x^4 \right) + \dfrac{d}{dx} \left(4x^2 \right) + \cancelto{0}{\dfrac{d}{dx}(1)} \\[8px] &= 4\dfrac{d}{dx}\left(x^4 \right) + 4\dfrac{d}{dx} \left(x^2 \right) \\[8px] &= 4\left(4 x^3 \right) + 4\left( 2x \right) \\[8px] &= 16x^3 + 8x \quad \cmark \end{align*} \]
[hide solution]

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II. Exponential Function Derivative
$$\bbox[yellow,5px]{\dfrac{d}{dx}e^x = e^x}$$ This one's easy to remember!
Exponential Differentiation Problem #1
Differentiate $f(x) = e^x + x$.
Show/Hide Solution
\[ \begin{align*} \dfrac{d}{dx}\left(e^x + x \right) &= \dfrac{d}{dx}\left(e^x \right) + \dfrac{d}{dx}(x) \\[8px] &= e^x + 1 \quad \cmark \end{align*} \]
[hide solution]
Exponential Differentiation Problem #2
Differentiate $f(x) = e^{1 + x}$.
Show/Hide Solution
Recall that $a^{a+b} = a^a a^b$. Hence $$e^{1 + x} = e \cdot e^x $$ And remember that $e$ is just a number — a constant. Hence it comes out in front of the derivative: \[ \begin{align*} \dfrac{d}{dx}\left(e^{1 + x} \right) &= \dfrac{d}{dx}\left(e \cdot e^x \right) \\[8px] &= e \dfrac{d}{dx}\left(e^x \right) \\[8px] &= e \left(e^x \right) \quad \cmark \\[8px] &= e^{1+x} \quad \cmark \end{align*} \] Note that the last two lines are completely equivalent. Either is correct as the answer.
[hide solution]
III. Trig Function Derivatives
\[ \bbox[yellow,5px]{ \begin{align*} \frac{d}{dx}\left(\sin x\right) &= \cos x &&& \frac{d}{dx}\left(\csc x\right) &= -\csc x \cot x \\ \\ \dfrac{d}{dx}\left(\cos x\right) &= -\sin x &&& \frac{d}{dx}\left(\sec x\right) &= \sec x \tan x \\ \\ \dfrac{d}{dx}\left(\tan x\right) &= \sec^2 x &&& \frac{d}{dx}\left(\cot x\right) &= -\csc^2 x \end{align*}} \] Notice that a negative sign appears in the derivatives of the co-functions: cosine, cosecant, and cotangent.
Trig Differentiation Problem #1
Differentiate $f(x) = \sin x - \cos x$.
Show/Hide Solution
Recall from the table that $\dfrac{d}{dx}(\sin x) = \cos x,$ and $\dfrac{d}{dx}(\cos x) = -\sin x.$ \[ \begin{align*} \dfrac{d}{dx} \left(\sin x – \cos x \right) &= \dfrac{d}{dx}(\sin x) – \dfrac{d}{dx}(\cos x) \\[8px] &= \cos x – (-\sin x) \\[8px] &= \cos x + \sin x \quad \cmark \end{align*} \]
[hide solution]
Trig Differentiation Problem #2
Differentiate $f(x) = 5x^3 - \tan x$.
Show/Hide Solution
Recall from the table that $\dfrac{d}{dx}(\tan x) = \sec^2 x.$ \[ \begin{align*} \dfrac{d}{dx}\left( 5x^3 – \tan x\right) &= 5\dfrac{d}{dx}\left(x^3 \right) – \dfrac{d}{dx}(\tan x) \\[8px] &= 5(3x^2) – \sec^2 x \\[8px] &= 15x^2 – \sec^2 x \quad \cmark \end{align*} \]
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IV. Product Rule
\[\bbox[yellow,5px]{ \begin{align*} \dfrac{d}{dx}(fg)&= \left(\dfrac{d}{dx}f \right)g + f\left(\dfrac{d}{dx}g \right)\\[8px] &= [{\small \text{ (derivative of the first) } \times \text{ (the second) }}]\, + \,[{\small \text{ (the first) } \times \text{ (derivative of the second)}}] \end{align*}}\]
Product Rule Problem #1
Differentiate $f(x) = x^3 e^x$.
Show/Hide Solution
Since the function is the product of two separate functions, $x^3$ and $e^x$, we must use the Product Rule: \[ \bbox[10px,border:2px dashed blue]{\begin{align*} \dfrac{d}{dx}(fg)&= \left(\dfrac{d}{dx}f \right)g + f\left(\dfrac{d}{dx}g \right)\\[8px] &= [{\small \text{ (derivative of the first) } \times \text{ (the second) }}]\, + \,[{\small \text{ (the first) } \times \text{ (derivative of the second)}}] \end{align*}}\] Recall that $\dfrac{d}{dx}x^3 = 3x^2,$ and that $\dfrac{d}{dx}e^x = e^x$: \begin{align*} \frac{d}{dx}\left(x^3 e^x \right) &= \left(\frac{d}{dx}x^3 \right) e^x + x^3 \left( \frac{d}{dx}e^x \right)\\[8px] &= 3x^2 e^x + x^3 e^x \quad \cmark \end{align*}
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Product Rule Problem #2
Differentiate $f(x) = x\sin x.$
\begin{array}{lll} \text{(A) }\cos x && \text{(B) }\sin x -x\cos x && \text{(C) } \sin x + x\cos x \end{array}
\begin{array}{ll}\text{(D) }-\cos x && \text{(E) None of these} \end{array}
Show/Hide Solution
Since the function is the product of two separate functions, $x$ and $\sin x$, we must use the Product Rule: Recall that $\dfrac{d}{dx}x = 1,$ and that $\dfrac{d}{dx}\sin x = \cos x$: \begin{align*} \dfrac{d}{dx} \left( x\sin x\right)&= \left(\dfrac{d}{dx}x\right)\sin x + x \left( \dfrac{d}{dx}\sin x \right) \\[8px] &= (1)\sin x + x \,(\cos x) \\[8px] &= \sin x + x\cos x \quad \implies \; \text{(C)}\;\cmark \end{align*}
[hide solution]
Product Rule Problem #3
Differentiate $g(\theta) = \sin \theta \, \cos \theta$.
Show/Hide Solution
Since the function is the product of two separate functions, $\sin \theta$ and $\cos \theta $, we must use the Product Rule: \[ \bbox[10px,border:2px dashed blue]{\begin{align*} \dfrac{d}{d\theta}(fg)&= \left(\dfrac{d}{d\theta}f \right)g + f\left(\dfrac{d}{d\theta}g \right)\\[8px] &= [{\small \text{ (derivative of the first) } \times \text{ (the second) }}]\, + \,[{\small \text{ (the first) } \times \text{ (derivative of the second)}}] \end{align*}}\] Recall that $\dfrac{d}{d \theta}\sin \theta = \cos \theta,$ and that $\dfrac{d}{d \theta}\cos \theta = -\sin \theta$: \begin{align*} \frac{d}{d\theta}(\sin \theta \, \cos \theta) &= \left(\frac{d}{d\theta} \sin \theta \right) \cos \theta + \sin \theta \left(\frac{d}{d\theta} \cos \theta \right) \\[8px] &= (\cos \theta) \cos \theta + \sin \theta (-\sin \theta) \\[8px] &= \cos^2 \theta – \sin^2 \theta \quad \cmark \end{align*}
[hide solution]
Product Rule Problem #4
Differentiate $f(x) = \left(e^x +1 \right) \tan x.$
Show/Hide Solution
Since the function is the product of two separate functions, $ \left(e^x +1 \right)$ and $\tan x$, we must use the Product Rule: \[ \bbox[10px,border:2px dashed blue]{ \begin{align*} \dfrac{d}{dx}(fg)&= \left(\dfrac{d}{dx}f \right)g + f\left(\dfrac{d}{dx}g \right)\\[8px] &= [{\small \text{ (derivative of the first) } \times \text{ (the second) }}]\, + \,[{\small \text{ (the first) } \times \text{ (derivative of the second)}}] \end{align*} }\] Recall that $\dfrac{d}{dx}\left(e^x + 1 \right) = e^x,$ and that $\dfrac{d}{dx}\tan x = \sec^2 x$: \[\begin{align*} \dfrac{d}{dx} \left[\left(e^x +1 \right) \tan x \right] &= \left[\dfrac{d}{dx} \left(e^x +1 \right) \right] \tan x + \left(e^x +1 \right)\left[ \dfrac{d}{dx}\tan x \right] \\[8px] &= \left[ e^x  \right]\tan x + \left(e^x +1 \right)\left[ \sec^2 x\right] \\[8px] &= e^x \tan x + \left(e^x +1 \right)\sec^2 x \quad \cmark \end{align*} \]
[hide solution]
Product Rule Problem #5
Differentiate $z(x) = x^{5/2} \, e^x \sin x$.
Show/Hide Solution
Since the function is the product of three separate functions, $x^{5/2}$, $e^x$, and $\sin x$, we must use an extension of the Product Rule: \[ \bbox[10px,border:2px dashed blue]{\begin{align*} \dfrac{d}{dx}(fgh)&= \left(\dfrac{d}{dx}f \right)gh + f\left(\dfrac{d}{dx}g \right)h + + fg\left(\dfrac{d}{dx}h \right)\\[8px] &= [{\small \text{ (derivative of the first) } \times \text{ (the second) }\times \text{ (the third) }}] \\ &\qquad + [{\small \text{ (the first) } \times \text{ (derivative of the second)}\times \text{ (the third) }}] \\ &\qquad \qquad + [{\small \text{ (the first) } \times\text{ (the second) } \times \text{ (derivative of the third)}}] \end{align*}}\] Recall that $\dfrac{d}{dx}x^{5/2} = \dfrac{5}{2}x^{3/2},$ and $\dfrac{d}{dx}e^x = e^x$, and that $\dfrac{d}{dx}\sin x = \cos x$. Then: \begin{align*} \dfrac{d}{dx}\left(x^{5/2} \, e^x \sin x \right) &= \left( \frac{d}{dx}x^{5/2}\right) e^x \sin x + x^{5/2} \left(\frac{d}{dx} e^x \right) \sin x + x^{5/2} e^x \left( \frac{d}{dx} \sin x \right) \\ \\ &= \left(\frac{5}{2}x^{3/2}\right) e^x \sin x + x^{5/2} \left(e^x \right) \sin x + x^{5/2} e^x (\cos x) \quad \cmark \\ \\ &= x^{3/2}e^x \left(\frac{5}{2}\sin x + x \sin x + x \cos x \right) \quad \cmark \end{align*} Note that the last two lines are equivalent and both are correct.
[hide solution]
Product Rule Problem #6
Given that $f(2) = 1$, $f'(2) = -3$, $g(2) = 4$, and $g'(2) = 8$, find $(fg)'(2)$.
Show/Hide Solution
Since we’re finding $(fg)’$, the derivative of the product of $f$ and $g$, we must use the Product Rule: \[ \bbox[10px,border:2px dashed blue]{\begin{align*} (fg’)&=f’g + fg’\\[8px] &= [{\small \text{ (derivative of the first) } \times \text{ (the second) }}]\, + \,[{\small \text{ (the first) } \times \text{ (derivative of the second)}}] \end{align*}}\] Hence: \begin{align*} (fg)'(2) &= f'(2) \cdot g(2) + f(2) \cdot g'(2) \\[8px] &= (-3)\cdot (4) + (1)\cdot (8) \\[8px] &= -12 + 8 = -4 \quad \cmark \end{align*}
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V. Quotient Rule
\[ \bbox[yellow,5px]{ \begin{align*} \dfrac{d}{dx}\left(\dfrac{f}{g} \right) &= \dfrac{\left(\dfrac{d}{dx}f \right)g - f\left(\dfrac{d}{dx}g \right)}{g^2} \\[8px] &=\dfrac{{[{\small \text{(derivative of the numerator) } \times \text{ (the denominator)}]}\\ \quad - \, [{\small \text{ (the numerator) } \times \text{ (derivative of the denominator)}}]}}{{\small \text{all divided by [the denominator, squared]}}} \end{align*}}\] Many students remember the quotient rule by thinking of the numerator as "hi," the demoninator as "lo," the derivative as "d," and then singing

"lo d-hi minus hi d-lo over lo-lo"

Quotient Rule Problem #1
Differentiate $f(x) = \dfrac{x^2}{e^x}$.
Show/Hide Solution
Since the function is the quotient of two separate functions, $x^2$ and $e^x$, we must use the Quotient Rule: \[ \bbox[10px,border:2px dashed blue]{ \begin{align*} \dfrac{d}{dx}\left(\dfrac{f}{g} \right) &= \dfrac{\left(\dfrac{d}{dx}f \right)g – f\left(\dfrac{d}{dx}g \right)}{g^2} \\[8px] &=\dfrac{{[{\small \text{(derivative of the numerator) } \times \text{ (the denominator)}]}\\ \quad – \, [{\small \text{ (the numerator) } \times \text{ (derivative of the denominator)}}]}}{{\small \text{all divided by [the denominator, squared]}}} \end{align*}}\] Recall that $\dfrac{d}{dx}x^2 = 2x$, and $\dfrac{d}{dx}e^x = e^x$. Hence: \begin{align*} \frac{d}{dx}\left(\frac{x^2}{e^x} \right) &= \frac{\left(\dfrac{d}{dx}x^2\right)e^x – x^2 \left(\dfrac{d}{dx}e^x \right)}{\left(e^x\right)^2} \\[8px] &= \frac{ \left( 2x \right) e^x – x^2 \left(e^x \right)}{\left(e^x\right)^2} \\[8px] &= \frac{e^x\left(2x – x^2 \right)}{\left(e^x\right)^2} \\[8px] &= \frac{2x – x^2 }{e^x} \quad \cmark \end{align*}
[hide solution]
Quotient Rule Problem #2
Differentiate $f(x) = \dfrac{\sin x}{x}.$
Show/Hide Solution
Since the function is the quotient of two separate functions, $\sin x$ and $x$, we must use the Quotient Rule: \[ \bbox[10px,border:2px dashed blue]{ \begin{align*} \dfrac{d}{dx}\left(\dfrac{f}{g} \right) &= \dfrac{\left(\dfrac{d}{dx}f \right)g – f\left(\dfrac{d}{dx}g \right)}{g^2} \\[8px] &=\dfrac{{[{\small \text{(derivative of the numerator) } \times \text{ (the denominator)}]}\\ \quad – \, [{\small \text{ (the numerator) } \times \text{ (derivative of the denominator)}}]}}{{\small \text{all divided by [the denominator, squared]}}} \end{align*}}\] Recall that $\dfrac{d}{dx}\sin x = \cos x,$ and $\dfrac{d}{dx}x = 1$: \[ \begin{align*} \dfrac{d}{dx}\left( \dfrac{\sin x}{x}\right) &= \frac{\left( \dfrac{d}{dx}\sin x \right)x – \sin x \left( \dfrac{d}{dx}x \right)}{x^2} \\[8px] &= \frac{(\cos x)x – \sin x \,(1)}{x^2} \\[8px] &= \frac{x \cos x – \sin x}{x^2} \quad \cmark \end{align*} \]
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Quotient Rule Problem #3
Differentiate $f(x) = \dfrac{e^x}{x+1}.$
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Since the function is the quotient of two separate functions, $e^x$ and $(x+1)$, we must use the Quotient Rule: \[ \bbox[10px,border:2px dashed blue]{ \begin{align*} \dfrac{d}{dx}\left(\dfrac{f}{g} \right) &= \dfrac{\left(\dfrac{d}{dx}f \right)g – f\left(\dfrac{d}{dx}g \right)}{g^2} \\[8px] &=\dfrac{{[{\small \text{(derivative of the numerator) } \times \text{ (the denominator)}]}\\ \quad – \, [{\small \text{ (the numerator) } \times \text{ (derivative of the denominator)}}]}}{{\small \text{all divided by [the denominator, squared]}}} \end{align*}}\] Recall that $\dfrac{d}{dx}e^x = e^x,$ and $\dfrac{d}{dx}(x+1) = 1$: \[ \begin{align*} \dfrac{d}{dx} \left(\frac{e^x}{x+1} \right) &= \frac{\left( \dfrac{d}{dx}e^x \right)(x+1) – e^x \left(\dfrac{d}{dx}(x+1) \right)}{(x+1)^2} \\[8px] &= \frac{\left( e^x \right)(x+1) – e^x(1)}{(x+1)^2} \\[8px] &= \frac{e^x(x+1 -1)}{(x+1)^2} \\[8px] &= \frac{xe^x}{(x+1)^2} \quad \cmark \end{align*} \]
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Quotient Rule Problem #4
Differentiate $f(x) = \dfrac{3x}{5 - \tan x}.$
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Since the function is the quotient of two separate functions, $3x$ and $(5 – \tan x)$, we must use the Quotient Rule: \[ \bbox[10px,border:2px dashed blue]{ \begin{align*} \dfrac{d}{dx}\left(\dfrac{f}{g} \right) &= \dfrac{\left(\dfrac{d}{dx}f \right)g – f\left(\dfrac{d}{dx}g \right)}{g^2} \\[8px] &=\dfrac{{[{\small \text{(derivative of the numerator) } \times \text{ (the denominator)}]}\\ \quad – \, [{\small \text{ (the numerator) } \times \text{ (derivative of the denominator)}}]}}{{\small \text{all divided by [the denominator, squared]}}} \end{align*}}\] Recall that $\dfrac{d}{dx}(3x) = 3,$ and $$\dfrac{d}{dx}(5 – \tan x) = \cancelto{0}{\dfrac{d}{dx}(5)} – \dfrac{d}{dx}\tan x = -\sec^2 x$$ Then \[ \begin{align*} \dfrac{d}{dx}\left(\frac{3x}{5 – \tan x} \right) &= \frac{\left(\dfrac{d}{dx}(3x) \right)(5 – \tan x) – (3x)\left(\dfrac{d}{dx}(5 – \tan x) \right)}{(5 – \tan x)^2} \\[8px] &= \frac{(3)(5 – \tan x) – (3x)(-\sec^2 x)}{(5 – \tan x)^2} \\[8px] &= \frac{15 – 3\tan x + 3x\sec^2 x}{(5 – \tan x)^2} \quad \cmark \end{align*} \]
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Quotient Rule Problem #5
Differentiate $g(u) = \dfrac{u^3 - 5u^2 + 6}{u -2}$.
Stop after taking the derivatives; don't bother to multiply out the terms and simplify.
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Since the function is the quotient of two separate functions, $u^3 – 5u^2 + 6$, and $u -2$, we must use the Quotient Rule: \[ \bbox[10px,border:2px dashed blue]{ \begin{align*} \dfrac{d}{du}\left(\dfrac{f}{g} \right) &= \dfrac{\left(\dfrac{d}{du}f \right)g – f\left(\dfrac{d}{du}g \right)}{g^2} \\[8px] &=\dfrac{{[{\small \text{(derivative of the numerator) } \times \text{ (the denominator)}]}\\ \quad – \, [{\small \text{ (the numerator) } \times \text{ (derivative of the denominator)}}]}}{{\small \text{all divided by [the denominator, squared]}}} \end{align*}}\] Recall that $\dfrac{d}{du}\left(u^3 – 5u^2 + 6 \right) = 3u^2 -10u$, and $\dfrac{d}{du}(u-2) = 1$. Hence \begin{align*} \frac{d}{du}\left(\frac{u^3 – 5u^2 + 6}{u -2} \right) &= \frac{\left(\dfrac{d}{du}(u^3 – 5u^2 + 6) \right)\left(u-2\right) – \left(u^3 – 5u^2 + 6 \right)\left( \dfrac{d}{du}(u-2)\right) }{\left(u -2 \right)^2} \\[8px] &= \frac{\left( 3u^2 -10u \right)(u-2) – (u^3 – 5u^2 + 6)(1) }{\left(u -2 \right)^2} \quad \cmark \end{align*}
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Quotient Rule Problem #6
Given that $f(2) = 1$, $f'(2) = -3$, $g(2) = 4$, and $g'(2) = 8$, find $\left(\dfrac{f}{g}\right)'(2)$.
Show/Hide Solution
Since we’re finding $\left(\dfrac{f}{g}\right)’$ the derivative of the quotient of $f$ over $g$, we must use the Quotient Rule: \[ \bbox[10px,border:2px dashed blue]{ \begin{align*} \dfrac{d}{dx}\left(\dfrac{f}{g} \right) &= \dfrac{\left(\dfrac{d}{dx}f \right)g – f\left(\dfrac{d}{dx}g \right)}{g^2} \\[8px] &=\dfrac{{[{\small \text{(derivative of the numerator) } \times \text{ (the denominator)}]}\\ \quad – \, [{\small \text{ (the numerator) } \times \text{ (derivative of the denominator)}}]}}{{\small \text{all divided by [the denominator, squared]}}} \end{align*}}\] Hence: \begin{align*} \left(\frac{f}{g}\right)'(2) &= \frac{f'(2) \cdot g(2) – f(2) \cdot g'(2)}{(g(2))^2} \\[8px] &= \frac{(-3)(4) – (1)(8)}{(4)^2} \\[8px] &= \frac{-12 – 8}{16} =- \frac{20}{16} = -\frac{5}{4} \quad \cmark \end{align*}
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More problems; University Exam Questions
The problems below mix ideas from above, and include questions from actual university exams.
Exam 1 Questions
Find the requested information.
(a) Let $f(u) = \dfrac{5u - 3}{u^2 + 1}$. Find $f'(u)$.
(b) Let $g(t) = (t^2 + 3 + t^{-2}) \tan t$. Find $g'(t)$.
Show/Hide Solution
Solution SummarySolution (a) DetailSolution (b) Detail
(a) $\dfrac{-5u^2 + 6u + 5}{(u^2+1)^2}$
(b) $ \left( 2t – 2t^{-3} \right) \tan t + (t^2 + 3 + t^{-2}) \sec^2 t$
\begin{align*} f'(u) &= \frac{d}{du} \left[\frac{5u – 3}{u^2 + 1} \right] \\ \\ &= \frac{\left( \frac{d}{du}(5u-3) \right) (u^2+1) – (5u -3) \left(\frac{d}{du}(u^2+1) \right)}{(u^2+1)^2} \\ \\ &= \frac{(5)(u^2 +1) – (5u -3)(2u)}{(u^2+1)^2} \\ \\ &= \frac{5u^2 + 5 – 10u^2 + 6u}{(u^2+1)^2} \\ \\ &= \frac{-5u^2 + 6u + 5}{(u^2+1)^2} \quad \cmark \end{align*}
\begin{align*} g'(t) &= \frac{d}{dt} \left[(t^2 + 3 + t^{-2}) \tan t \right] \\ \\ &= \left( \frac{d}{dt} (t^2 + 3 + t^{-2}) \right) \tan t + (t^2 + 3 + t^{-2}) \left( \frac{d}{dt} \tan t \right) \\ \\ &= \left( 2t – 2t^{-3} \right) \tan t + (t^2 + 3 + t^{-2}) \sec^2 t \quad \cmark \end{align*}
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VERY Common Exam Question: Find a & b (#1)
Note: This type of question frequently appears on exams, and we strongly urge you to learn how to solve it.

Find all $a$ and $b$ such that the function $f(x)$ is differentiable: $$ f(x) = \begin{cases} ax + b & \text{if } x \lt 1 \\ x^4 + x +1 & \text{if } x \geq 1 \end{cases} $$
Show/Hide Solution
For $f(x)$ to be differentiable at $x=1$, at it must (I) be continuous there, and (II) have the same derivative whether we approach from the left or from the right. These two conditions will allow us to solve for $a$ and $b$.

(I) $f(x)$ must be continuous at $x = 1$, which means $$\lim_{x \to 1^-}f(x) = \lim_{x \to 1^+}f(x) $$ Now, from the left, we look at the function given for $x < 1$: $$\lim_{x \to 1^-}f(x) = \lim_{x \to 1^-}(ax + b) = a + b $$ While from the right, we look at the function for given for $x > 1$: $$\lim_{x \to 1^+}f(x) = \lim_{x \to 1^+}(x^4 + x +1) = 1 + 1 +1 = 3 $$ Hence for $f$ to be continuous, we must have \begin{align*} \lim_{x \to 1^-}f(x) &= \lim_{x \to 1^+}f(x) \\[4px] a+b &= 3 \end{align*}

(II) $f(x)$ must have the same derivative whether we approach $x=1$ from the left or the right: $$\lim_{x \to 1^-}f'(x) = \lim_{x \to 1^+}f'(x) $$ From the left: $$\lim_{x \to 1^-}f'(x) = \lim_{x \to 1^-}\left[\frac{d}{dx}(ax+b) \right]= \lim_{x \to 1^-}a = a $$ While from the right: $$\lim_{x \to 1^+}f'(x) = \lim_{x \to 1^+}\left[\frac{d}{dx}(x^4 + x +1) \right]= \lim_{x \to 1^+}(4x^3 + 1) = 5 $$ Hence we must have \begin{align*} \lim_{x \to 1^-}f(x) &= \lim_{x \to 1^+}f(x) \\[4px] a &= 5 \end{align*}

And since our equation above says $a + b = 3$, if $a = 5$ then $b =-2$. $\quad \cmark$
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VERY Common Exam Question: Find a & b (#2)
Note: This type of question frequently appears on exams, and we strongly urge you to learn how to solve it.

Find the values of $a$ and $b$ that will make the function $f(x)$ differentiable. $$ f(x) =\begin{cases} ax + b & \text{if } x \lt \pi \\ \sin x & \text{if } x \geq \pi \end{cases} $$
Show/Hide Solution
For $f(x)$ to be differentiable at $x=\pi$, at it must (I) be continuous there, and (II) have the same derivative whether we approach from the left or from the right. These two conditions will allow us to solve for $a$ and $b$.

(I) $f(x)$ must be continuous at $x = \pi$: $$\lim_{x \to \pi^-}f(x) = \lim_{x \to \pi^+}f(x) $$ Now, from the left, we look at the function given for $x < \pi$: $$ \lim_{x \to \pi^-}f(x) = \lim_{x \to \pi^-}(ax + b) = a\pi + b $$ While from the right, we look at the function given for $x > \pi$: $$ \lim_{x \to \pi^+}f(x) = \lim_{x \to \pi^+} \sin x = 0 $$ Hence we must have \begin{align*} \lim_{x \to \pi^-}f(x) &= \lim_{x \to \pi^+}f(x) \\[4px] a\pi + b &= 0 \end{align*}

(II) $f(x)$ must have the same derivative whether we approach $x = \pi$ from the left or the right: $$\lim_{x \to \pi^-}f'(x) = \lim_{x \to \pi^+}f'(x) $$ From the left: $$\lim_{x \to \pi^-}f'(x) = \lim_{x \to \pi^-}\left[\frac{d}{dx}(ax+b) \right]= \lim_{x \to \pi^-}a = a $$ While from the right: $$\lim_{x \to \pi^+}f'(x) = \lim_{x \to \pi^+}\left[\frac{d}{dx}(\sin x) \right]= \lim_{x \to \pi^+}\cos x = -1$$ Hence we must have \begin{align*} \lim_{x \to \pi^-}f'(x) &= \lim_{x \to \pi^+}f'(x) \\[4px] a &= -1 \end{align*}
And since our equation above says $a\pi + b = 0$, if $a = -1$ then $b = \pi$. $\quad \cmark$
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Table of Values
Let $f$ and $g$ be differentiable functions and let the values of $f, g, f'$ and $g'$ at $x=1$ and $x=2$ be given by the table.

\begin{array}{c | c | c | c | c} x & f(x) & g(x) & f'(x) & g'(x)\\ \hline 1 & 5 & 3 & 2 & 7 \\ \hline 2 & -2 & 1 & 4 & 6 \\ \end{array}
Find the numerical value of each of the following:
(a) the derivative of $\displaystyle{\frac{f}{g}}$ at $x = 1$
(b) $\displaystyle{\lim_{h \to 0} \frac{9 \cdot f(1+h) - 9 \cdot f(1)}{h}}$
(c) $\displaystyle{\lim_{h \to 0} \frac{f(2+h) \cdot g(2+h) - f(2) \cdot g(2)}{h}}$
(d) $\displaystyle{\lim_{h \to 0} \dfrac{g(2) - g(2+h)}{h}}$
Show/Hide Solution
Solution SummarySolution (a) DetailSolution (b) DetailSolution (c) DetailSolution (d) Detail
(a) $-\dfrac{29}{9}$
(b) 18
(c) $-8$
(d) $-6$
We use the quotient rule: \begin{align*} \frac{d}{dx}\left[ \frac{f}{g} \right]_{\text{at }x=1} &= \left[ \frac{f’g – fg’}{g^2} \right]_{\text{at }x=1} \\ \\ &= \frac{(2)(3) – (5)(7)}{3^2} \\ \\ &= -\frac{29}{9} \quad \cmark \end{align*}
The key here is to recognize that if we factor out the 9, we are left with the definition of the derivative of $f$ at $x=1$: $$\lim_{h \to 0} \frac{f(1+h) – f(1)}{h} = f'(1)$$ Hence \begin{align*} \lim_{h \to 0} \frac{9 \cdot f(1+h) – 9 \cdot f(1)}{h} &= \lim_{h \to 0}9 \cdot \lim_{h \to 0} \frac{f(1+h) – f(1)}{h} \\ \\ &= 9 \cdot f'(1) \\ \\ &= 9 \cdot 2 = 18 \quad \cmark \end{align*}
The key here is to recognize this quantity as the definition of the derivative for the product of $f$ and $g$ at $x=2$:

\begin{align*} \lim_{h \to 0} \frac{f(2+h) \cdot g(2+h) – f(2) \cdot g(2)}{h} &= \frac{d}{dx} \left[f(x)g(x) \right]_{\text{at }x=2}\\ \\ &= f'(2)g(2) + f(2)g'(2) &&\text{ [by the product rule]}\\ \\ &= (4)(1) + (-2)(6)\\ \\ &= -8 \quad \cmark \end{align*}
This quantity reminds us of the derivative of $g$ at $x=2$, but seems “backwards.”
Factoring out a $-1$ makes apparent the definition we’re used to seeing:

\begin{align*} \lim_{h \to 0} \frac{g(2) – g(2+h)}{h} &= (-1) \lim_{h \to 0} \frac{g(2+h) – g(2)}{h} \\ \\ &= (-1) g'(2) \\ \\ &= (-1)(6) = -6 \quad \cmark \end{align*}
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