Calculate Derivatives – Problems & Solutions

This screen contains only a summary of the rules needed to quickly calculate derivatives, and free problems for you to practice, each with a complete solution immediately available. If you need to quickly get this down, and don't have time to work through the more comprehensive material earlier in this Chapter, this screen is for you.

And if you're looking to review and practice for an exam, we suggest going to the next page on Chain Rule problems, since those are more advanced and encompass the use of all of the other rules.

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 derivatives] [Product rule] [Quotient rule] [Mixed problems] [Chain rule]

CALCULUS SUMMARY: Derivatives and Rules

Let's step through each of the rules and give you the chance to try a few problems. Everything comes together, including more challenging problems, on the next screen regarding the Chain Rule.

I. Power Rule, $𝑥 𝑛$

Most frequently, you will use the Power Rule:

𝑑 𝑑 𝑥 (𝑥 𝑛) = 𝑛 𝑥 𝑛 - 1

Power RulePractice Problem 1

Differentiate

𝑓 (𝑥) = 2 𝜋

Power Rule Practice Problem 2

Differentiate

𝑓 (𝑥) = 23 𝑥 9

Power Rule Practice Problem 3

Differentiate

𝑓 (𝑥) = 2 𝑥 3 - 4 𝑥 2 + 𝑥 - 33

Power Rule Practice Problem 4

Differentiate

𝑓 (𝑥) = 𝑥 1001 + 5 𝑥 3 - 6 𝑥 + 10, 687

Power Rule Practice Problem 5

Show that

𝑑 𝑑 𝑥 \sqrt 𝑥 = 12 1 \sqrt 𝑥

.
Recall that

\sqrt 𝑥 = 𝑥 1 / 2 .

Power Rule Practice Problem 6

Differentiate

𝑓 (𝑥) = 5 𝑥 3

.
Recall that

1 𝑥 𝑛 = 𝑥 - 𝑛 .

Power Rule Practice Problem 7

Differentiate

𝑓 (𝑥) = 1 𝑥 2 - 4 𝑥 5

Power Rule Practice Problem 8

Differentiate

𝑓 (𝑥) = 3 \sqrt 𝑥 - 1 \sqrt 𝑥

Power Rule Practice Problem 9

Differentiate

𝑓 (𝑥) = \sqrt 𝑥 (𝑥 2 - 8 + 1 𝑥)

Power Rule Practice Problem 10

Differentiate

𝑓 (𝑥) = (2 𝑥 2 + 1) 2

II. Exponential, $𝑒 𝑥$

𝑑 𝑑 𝑥 𝑒 𝑥 = 𝑒 𝑥

This one's easy to remember!

Exponential Practice Problem 1

Differentiate

𝑓 (𝑥) = 𝑒 𝑥 + 𝑥

Exponential Practice Problem 2

Differentiate

𝑓 (𝑥) = 𝑒 1 + 𝑥

More problems with exponentials are below in the Product Rule and Quotient Rule sections. And really, most problems involving the derivative of an exponential also require the Chain Rule, and those problems are on the next screen.

III. Trig Function Derivatives

𝑑 𝑑 𝑥 (s i n 𝑥) = c o s 𝑥 𝑑 𝑑 𝑥 (c s c 𝑥) = - c s c 𝑥 c o t 𝑥 𝑑 𝑑 𝑥 (c o s 𝑥) = - s i n 𝑥 𝑑 𝑑 𝑥 (s e c 𝑥) = s e c 𝑥 t a n 𝑥 𝑑 𝑑 𝑥 (t a n 𝑥) = s e c 2 𝑥 𝑑 𝑑 𝑥 (c o t 𝑥) = - c s c 2 𝑥

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

Trig Function Practice Problem 1

Differentiate

𝑓 (𝑥) = s i n 𝑥 - c o s 𝑥

Trig Function Practice Problem 2

Differentiate

𝑓 (𝑥) = 5 𝑥 3 - t a n 𝑥

More problems with trig functions are below in the Product Rule and Quotient Rule sections. And then of course most problems involving the derivative of a trig function also requirethe Chain Rule so more problems are there.

IV. Product Rule

The Product Rule is used to find the derivative of the product of two functions:

𝑑 𝑑 𝑥 (𝑓 𝑔) = (𝑑 𝑑 𝑥 𝑓) 𝑔 + 𝑓 (𝑑 𝑑 𝑥 𝑔) = [(d e r i v a t i v e o f t h e f i r s t) \times (t h e s e c o n d)] + [(t h e f i r s t) \times (d e r i v a t i v e o f t h e s e c o n d)]

Product Rule Practice Problem 1

Differentiate

𝑓 (𝑥) = 𝑥 3 𝑒 𝑥

Product Rule Practice Problem 3

Differentiate

𝑔 (𝜃) = s i n 𝜃 c o s 𝜃

Product Rule Practice Problem 4

Differentiate

𝑓 (𝑥) = (𝑒 𝑥 + 1) t a n 𝑥 .

Product Rule Practice Problem 5

Differentiate

𝑧 (𝑥) = 𝑥 5 / 2 𝑒 𝑥 s i n 𝑥

Product Rule Practice Problem 6

Given that

𝑓 (2) = 1

𝑓' (2) = - 3

𝑔 (2) = 4

, and

𝑔' (2) = 8

, find

(𝑓 𝑔)' (2)

V. Quotient Rule

The Quotient Rule is used to find the derivative of the quotient of two functions:

𝑑 𝑑 𝑥 (𝑓 𝑔) = (𝑑 𝑑 𝑥 𝑓) 𝑔 - 𝑓 (𝑑 𝑑 𝑥 𝑔) 𝑔 2 = [(d e r i v a t i v e o f t h e n u m e r a t o r) \times (t h e d e n o m i n a t o r)] - [(t h e n u m e r a t o r) \times (d e r i v a t i v e o f t h e d e n o m i n a t o r)] a l l d i v i d e d b y [t h e d e n o m i n a t o r, s q u a r e d]

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 Practice Problem 1

Differentiate

𝑓 (𝑥) = 𝑥 2 𝑒 𝑥

Quotient Rule Practice Problem 2

Differentiate

𝑓 (𝑥) = s i n 𝑥 𝑥 .

Quotient Rule Practice Problem 3

Differentiate

𝑓 (𝑥) = 𝑒 𝑥 𝑥 + 1 .

Quotient Rule Practice Problem 4

Differentiate

𝑓 (𝑥) = 3 𝑥 5 - t a n 𝑥 .

Quotient Rule Practice Problem 5

Differentiate

𝑔 (𝑢) = 𝑢 3 - 5 𝑢 2 + 6 𝑢 - 2

.
Stop after taking the derivatives; don't bother to multiply out the terms and simplify.

Quotient Rule Practice Problem 6

Given that

𝑓 (2) = 1

𝑓' (2) = - 3

𝑔 (2) = 4

, and

𝑔' (2) = 8

, find

(𝑓 𝑔)' (2)

VI. Mixed Problems

The problems below mix ideas from above, and include questions from actual university exams.

Mixed Problem 1

Find the requested information.

(a)

Let

𝑓 (𝑢) = 5 𝑢 - 3 𝑢 2 + 1

. Find

𝑓' (𝑢)

(b)

Let

𝑔 (𝑡) = (𝑡 2 + 3 + 𝑡 - 2) t a n 𝑡

. Find

𝑔' (𝑡)

$Tip icon$

Note: Problems similar to the following frequently appear on exams, and so we strongly urge you to learn how to solve it.

Mixed Problem 2

Find the values of

𝑎

and

𝑏

that will make the function

𝑓 (𝑥)

differentiable.

𝑓 (𝑥) = {𝑎 𝑥 + 𝑏 i f 𝑥 < 𝜋 s i n 𝑥 i f 𝑥 \geq 𝜋

For

𝑓 (𝑥)

to be differentiable at

𝑥 = 𝜋

, 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 = 𝜋

l i m 𝑥 \to 𝜋 - 𝑓 (𝑥) = l i m 𝑥 \to 𝜋 + 𝑓 (𝑥)

Now, from the left, we look at the function given for

𝑥 < 𝜋

l i m 𝑥 \to 𝜋 - 𝑓 (𝑥) = l i m 𝑥 \to 𝜋 - (𝑎 𝑥 + 𝑏) = 𝑎 𝜋 + 𝑏

While from the right, we look at the function given for

𝑥 > 𝜋

l i m 𝑥 \to 𝜋 + 𝑓 (𝑥) = l i m 𝑥 \to 𝜋 + s i n 𝑥 = 0

Hence we must have

l i m 𝑥 \to 𝜋 - 𝑓 (𝑥) = l i m 𝑥 \to 𝜋 + 𝑓 (𝑥) 𝑎 𝜋 + 𝑏 = 0

(II)

𝑓 (𝑥)

must have the same derivative whether we approach

𝑥 = 𝜋

from the left or the right:

l i m 𝑥 \to 𝜋 - 𝑓' (𝑥) = l i m 𝑥 \to 𝜋 + 𝑓' (𝑥)

From the left:

l i m 𝑥 \to 𝜋 - 𝑓' (𝑥) = l i m 𝑥 \to 𝜋 - [𝑑 𝑑 𝑥 (𝑎 𝑥 + 𝑏)] = l i m 𝑥 \to 𝜋 - 𝑎 = 𝑎

While from the right:

l i m 𝑥 \to 𝜋 + 𝑓' (𝑥) = l i m 𝑥 \to 𝜋 + [𝑑 𝑑 𝑥 (s i n 𝑥)] = l i m 𝑥 \to 𝜋 + c o s 𝑥 = - 1

Hence we must have

l i m 𝑥 \to 𝜋 - 𝑓' (𝑥) = l i m 𝑥 \to 𝜋 + 𝑓' (𝑥) 𝑎 = - 1

And since our equation above says

𝑎 𝜋 + 𝑏 = 0

, if

𝑎 = - 1

then

𝑏 = 𝜋

✓

[close solution]

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Calculate Derivatives – Problems & Solutions

$𝑥 𝑛$ Derivatives

Exponential Derivative

Trigonometric Derivatives

Constant Factor Rule

Sum of Functions Rule

Product Rule for Derivatives

IV. Quotient Rule for Derivatives

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Calculate Derivatives – Problems & Solutions

𝑥𝑛 Derivatives

Exponential Derivative

Trigonometric Derivatives

Constant Factor Rule

Sum of Functions Rule

Product Rule for Derivatives

IV. Quotient Rule for Derivatives

How helpful was this page?

$𝑥 𝑛$ Derivatives