We’ll solve this two ways. The first is the way most experienced people quickly develop the answer, and that we hope you’ll soon be comfortable with. The second is more formal. Solution 1 (quick, the way most people reason). Think something like: “The function is some stuff to the eighth-power. So the derivative is eight times that same stuff to the seventh power, times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}\] \begin{align*} f(x) &= (\text{stuff})^8; \quad \text{stuff} = 3x^2 – 4x + 5 \\[12px] \text{Then}\phantom{f(x)= }\\ \dfrac{df}{dx} &= 8(\text{stuff})^7 \cdot \dfrac{d}{dx}\left(3x^2 – 4x + 5\right) \\[8px] &= 8\left(3x^2 – 4x + 5\right)^7 \cdot (6x -4) \implies \text{ (B)} \quad \cmark \end{align*} Note: You’d never actually write out “stuff = ….” Instead just hold in your head what that “stuff” is, and proceed to write down the required derivatives. Solution 2 (more formal). Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]
&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] We have the outer function $f(u) = u^8$ and the inner function $u = g(x) = 3x^2 – 4x + 5.$ Then $f'(u) = 8u^7,$ and $g'(x) = 6x -4.$ Hence \begin{align*} f'(x) &= 8u^7 \cdot (6x – 4) \\[8px] &= 8\left(3x^2 – 4x + 5\right)^7 \cdot (6x-4) \implies \text{ (B)} \quad \cmark \end{align*}

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We’ll again solve this two ways. The first is the way most experienced people quickly develop the answer, and that we hope you’ll soon be comfortable with. The second is more formal. Solution 1 (quick, the way most people reason). Think something like: “The function is some stuff to the power of 3. So the derivative is 3 times that same stuff to the power of 2, times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}\] \begin{align*} f(x) &= \big[\text{stuff}\big]^3; \quad \text{stuff} = \tan x \\[12px] \text{Then}\phantom{f(x)= }\\ \dfrac{df}{dx} &= 3\big[\text{stuff}\big]^2 \cdot \dfrac{d}{dx}(\tan x) \\[8px] &= 3\big[\tan x\big]^2 \cdot \sec^2 x \\[8px] &= 3\tan^2 x \cdot \sec^2 x \implies \text{ (D)} \quad \cmark \end{align*} Note: You’d never actually write out “stuff = ….” Instead just hold in your head what that “stuff” is, and proceed to write down the required derivatives. Solution 2 (more formal) . Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]
&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]
&\qquad \times \text{ [derivative of the inner function]}
\end{align*}}\] We have the outer function $f(u) = u^3$ and the inner function $u = g(x) = \tan x.$ Then $f'(u) = 3u^2,$ and $g'(x) = \sec^2 x.$ (Recall that $(\tan x)’ = \sec^2 x.$) Hence \begin{align*} f'(x) &= 3u^2 \cdot (\sec^2 x) \\[8px] &= 3\big[\tan x\big]^2 \cdot \sec^2 x \\[8px] &= 3\tan^2 x \cdot \sec^2 x \implies \text{ (D)} \quad \cmark \\[8px] \end{align*}

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Solution 1. Think something like: “The function is $\sqrt{\text{stuff}}$. So the derivative is $\dfrac{1}{2}\dfrac{1}{\sqrt{\text{that same stuff}}}$, times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}\] \begin{align*} f(x) &= \sqrt{\text{stuff}}; \quad \text{stuff} = x^2 + 1 \\[12px] \text{Then}\phantom{f(x)= }\\ \dfrac{df}{dx} &= \frac{1}{2}\frac{1}{\sqrt{\text{stuff}}} \cdot \left(\frac{d}{dx}(\text{stuff})\right) \\[8px] &= \frac{1}{2}\frac{1}{\sqrt{{x^2 + 1}}}\cdot ( 2x ) \implies \text{ (A)} \quad \cmark \end{align*} We could of course cancel the 2’s, but we’re leaving the result as-is so you can easily see how we applied the Chain rule. Solution 2. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]
&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]
&\qquad \times \text{ [derivative of the inner function]}
\end{align*}}\] We have the outer function $f(u) = \sqrt{u}$ and the inner function $u = g(x) = x^2 + 1.$ Then $\left(\sqrt{u} \right)’ = \dfrac{1}{2}\dfrac{1}{ \sqrt{u}},$ and $\left(x^2 + 1 \right)’ = 2x.$ Hence \begin{align*} f'(x) &= \dfrac{1}{2}\dfrac{1}{ \sqrt{u}} \cdot 2x \\[8px] &= \dfrac{1}{2}\dfrac{1}{ \sqrt{x^2+1}} \cdot 2x \implies \text{ (A)} \quad \cmark \end{align*}

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Solution 1 (quick, the way most people reason). Think something like: “The function is $\sqrt{\text{stuff}}$. So the derivative is $\dfrac{1}{2}\dfrac{1}{\sqrt{\text{that same stuff}}}$, times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}\] \begin{align*} f(x) &= \sqrt{\text{stuff}}; \quad \text{stuff} = \sin x \\[12px] \text{Then}\phantom{f(x)= }\\ \dfrac{df}{dx} &= \frac{1}{2}\frac{1}{\sqrt{\text{stuff}}} \cdot \left(\frac{d}{dx}(\text{stuff})\right) \\[8px] &= \frac{1}{2}\frac{1}{\sqrt{\sin x}}\cdot \cos x \implies \text{ (C)} \quad \cmark \end{align*} Solution 2 (more formal). Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]
&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]
&\qquad \times \text{ [derivative of the inner function]}
\end{align*}}\] We have the outer function $f(u) = \sqrt{u}$ and the inner function $u = g(x) = \sin x.$ Then $\left(\sqrt{u} \right)’ = \dfrac{1}{2}\dfrac{1}{ \sqrt{u}},$ and $(\sin x)’ = \cos x.$ Hence \begin{align*} f'(x) &= \dfrac{1}{2}\dfrac{1}{ \sqrt{u}} \cdot \cos x \\[8px] &= \dfrac{1}{2}\dfrac{1}{ \sqrt{\sin x}} \cdot \cos x \implies \text{ (C)} \quad \cmark \end{align*}

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Solution 1 (quick, the way most people reason). Think something like: “The function is $e$ to the power of some stuff. So the derivative is $e$ to the power of exactly the same stuff, times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}\] \begin{align*} f(x) &= e^{\text{stuff}}; \quad \text{stuff} = \sin x \\[12px] \text{Then}\phantom{f(x)= }\\ \dfrac{df}{dx} &= \left(e^{\text{stuff}} \right) \left(\dfrac{d}{dx}\text{(stuff)} \right) \\[8px] &= e^{\sin x} \cos x \implies \text{ (D)} \quad \cmark \\[8px] \end{align*} Note: You’d never actually write out “stuff = ….” Instead just hold in your head what that “stuff” is, and proceed to write down the required derivatives. Solution 2 (more formal). Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]
&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]
&\qquad \times \text{ [derivative of the inner function]}
\end{align*}}\] We have the outer function $f(u) = e^u$ and the inner function $u = g(x) = \sin x.$ Then $f'(u) = e^u,$ and $g'(x) = \cos x.$ Hence \begin{align*} f'(x) &= e^u \cdot \cos x \\[8px] &= e^{\sin x} \cdot \cos x \implies \text{ (D)} \quad \cmark \end{align*}

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Solution 1 (quick, the way most people reason). Think something like: “The function is $e$ to the power of some stuff. So the derivative is $e$ to the power of exactly the same stuff, times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}\] \begin{align*} f(x) &= e^{\text{stuff}}; \quad \text{stuff} = x^2 \\[12px] \text{Then}\phantom{f(x)= }\\ \dfrac{df}{dx} &= \left(e^{\text{stuff}} \right) \left(\dfrac{d}{dx}\text{(stuff)} \right) \\[8px] &= e^{x^2} \cdot 2x \implies \text{ (B)} \quad \cmark \\[8px] \end{align*} Solution 2 (more formal). Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]
&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]
&\qquad \times \text{ [derivative of the inner function]}
\end{align*}}\] We have the outer function $f(u) = e^u$ and the inner function $u = g(x) = x^2.$ Then $f'(u) = e^u,$ and $g'(x) = 2x.$ Hence \begin{align*} f'(x) &= e^u \cdot 2x \\[8px] &= e^{x^2} \cdot 2x \implies \text{ (B)} \quad \cmark \end{align*}

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Solution 1. Think something like: “The function is sin(of some stuff). So the derivative is cos(of that same stuff), times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}\] \begin{align*} f(x) &= \sin(\text{stuff}); \quad \text{stuff} = 2x \\[12px] \text{Then}\phantom{f(x)= }\\ \dfrac{df}{dx} &= \cos(\text{stuff}) \left(\dfrac{d}{dx}(\text{stuff}) \right) \\[4px] &= \cos(2x) \cdot(2) \implies \text{ (A)} \quad \cmark \end{align*} Solution 2. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]
&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]
&\qquad \times \text{ [derivative of the inner function]}
\end{align*}}\] We have the outer function $f(u) = \sin u$ and the inner function $u = g(x) = 2x.$ Then $f'(u) = \cos u,$ and $g'(x) = 2.$ Hence \begin{align*} f'(x) &= \cos u \cdot 2 \\[8px] &= \cos(2x) \cdot 2 \implies \text{ (A)} \quad \cmark \end{align*}

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Solution 1. Think something like: “The function is tan(of some stuff). So the derivative is $\sec^2 \text{(of that same stuff)}$, times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}\] \begin{align*} f(x) &= \tan(\text{stuff}); \quad \text{stuff} = e^x \\[12px] \text{Then}\phantom{f(x)= }\\ \dfrac{df}{dx} &= \sec^2(\text{stuff}) \left(\dfrac{d}{dx}(\text{stuff}) \right) \\[4px] &= \sec^2(e^x) \cdot e^x \implies \text{ (D)} \quad \cmark \end{align*} Solution 2. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]
&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]
&\qquad \times \text{ [derivative of the inner function]}
\end{align*}}\] We have the outer function $f(u) = \tan u$ and the inner function $u = g(x) = e^x.$ Then $f'(u) = \sec^2 u,$ and $g'(x) = e^x.$ Hence \begin{align*} f'(x) &= \sec^2 u \cdot e^x \\[8px] &= \sec^2(e^x) \cdot e^x \implies \text{ (D)} \quad \cmark \end{align*}

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Let’s first think about the derivative of each term separately. We won’t write out “stuff” as we did before to use the Chain Rule, and instead will just write down the answer using the same thinking as above: $$\left[ \left(x^2 + 1 \right)^7\right]’ = 7\left(x^2 + 1 \right)^6 \cdot (2x) $$ and $$\left[(3x – 7)^4 \right]’ = 4(3x – 7)^3 \cdot (3) $$ Now let’s use the Product Rule: \[ \begin{align*} [f(x)\,g(x)]’ &= f'(x)\, g(x)\qquad +\qquad f(x)\,g'(x) \\[8px] \left[\left(x^2 + 1 \right)^7 (3x – 7)^4 \right]’ &= \left[ \left(x^2 + 1 \right)^7\right]’ (3x – 7)^4\, + \,\left(x^2 + 1 \right)^7 \left[(3x – 7)^4 \right]’ \\[8px] &= \left[7\left(x^2 + 1 \right)^6 \cdot (2x) \right](3x – 7)^4 + \left(x^2 + 1 \right)^7 \left[4(3x – 7)^3 \cdot (3) \right] \implies \text{ (C)} \quad \cmark \end{align*} \] We could of course simplify this expression algebraically: $$f'(x) = 14x\left(x^2 + 1 \right)^6 (3x – 7)^4 + 12 \left(x^2 + 1 \right)^7 (3x – 7)^3 $$ We instead stopped where we did above to emphasize the way we’ve developed the result, which is what matters most here. (You don’t need us to show you how to do algebra!) Besides, on an exam your grader is most likely to check for something that looks like our result, which shows that you know how to use both the Product and Chain Rules correctly. You might ask your teacher how much you should simplify on an exam: small algebraic mistakes often happen during the “simplification process,” so if you won’t lose points by not simplifying at all, that’s your best bet.

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Let’s first think about the derivative of the numerator (we’ll call $f(x)$) and the denominator (we’ll call $g(x)$) separately. First, the numerator: \begin{align*} f(x) &= e^{2x} \\[4px] f'(x) &= e^{2x}\cdot 2 \end{align*} And now the denominator: \begin{align*} g(x) &= 1-x^2 \\[4px] g'(x) &= -2x \end{align*} Now let’s use the Quotient rule: \begin{align*} \left[\dfrac{f(x)}{g(x)} \right]’ &= \dfrac{f'(x)\,g(x) – f(x)\,g'(x)}{\left[g(x) \right]^2} \\[8px] \left[ \dfrac{e^{2x}}{1-x^2} \right]’ &= \dfrac{\left( e^{2x}\cdot 2\right)\left(1-x^2 \right) – \left(e^{2x} \right)\left(-2x \right)}{\left(1-x^2 \right)^2} \implies \text{ (B)} \quad \cmark \end{align*} We could simplify the answer, but prefer to leave it as-written to keep the focus on how the Chain rule simply enters in when you compute $f'(x)$ and $g'(x).$ Otherwise, you’re just using the Quotient rule as you did earlier.

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\begin{align*} \left[ f\big(g(x)\big)\right]’ &= f’\big(g(x)\big) \cdot g'(x) \\[5px] &=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px] &\qquad \times \text{ [derivative of the inner function]} \end{align*} \begin{align*} \left[ f\big(g(2)\big)\right]’ &= f’\big(g(2)\big) \cdot \left(g'(2)\right) \\ \\ &= f'(4) \cdot (8) \\ \\ &= 5 \cdot 8 = 40 \implies \text{ (C)} \quad \cmark \end{align*}

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Using the Chain Rule at $x=0$:
\[\left. \Big[ f\Big(g(x)\Big) \Big]’\right._{x=0} = f’\Big( g(0) \Big) \cdot g'(0)\] We’ll break this into parts: (1) We’ll first find $f’\Big( g(0) \Big),$ and then (2) we’ll find $g'(0).$

Subproblem 1: Find $f’\Big( g(0) \Big).$
We first need to determine $g(0),$ which we can read from the graph: $g(0) = 2.$ Hence $f’\Big( g(0) \Big) = f'(2).$
This value equals the slope of the line segment for $f$ that contains $x=2$.
You could use any two points on that line segment to find the slope value. We’ll use the convenient points $(2,1)$ and $(3,-2)$: \[f'(-2)= \dfrac{-2-1}{3-2} = \dfrac{-3}{1} = -3 \] Subproblem 2: Find $g'(0).$
Similarly, $g'(0)$ equals the slope of the line segment for $g$ that contains $x=0,$ which is zero since that line segment is horizontal: \[g'(0) = 0\] Then substituting these values $f’\Big( g(0) \Big) = f'(2) = -3$ and $g'(0) = 0$ into the Chain Rule: \begin{align*} \left. \Big[ f\Big(g(x)\Big) \Big]’\right._{x=0} &= f’\Big( g(0) \Big) \cdot g'(0) \\[8px] &= (-3)\cdot(0) \\[8px] &= 0 \implies \quad \text{ (A) } \quad \cmark \end{align*}

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The key realization in this problem is that the requested limit is the definition of the derivative of $f(g(x)),$ $\left[f(g(x)) \right]’,$ at $x=2$: \[ \lim_{h \to 0} \frac{f(g(2+h)) – f(g(2))}{h} = \frac{d}{dx}\left[f(g(x)) \right]_{\text{at }x=2}\] Once we realize that, we can use what we know about the Chain rule and read the necessary values from the table: \begin{align*} \lim_{h \to 0} \frac{f(g(2+h)) – f(g(2))}{h} &= \frac{d}{dx}\left[f(g(x)) \right]_{\text{at }x=2} \\[8px] &= f'(g(2)) \cdot g'(2) &&\text{ [by the chain rule]} \\[8px] &= f'(1) \cdot g'(2) &&[g(2) = 1]\\[8px] &= (2)(6) = 12 \quad \cmark \end{align*}

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\begin{align*} f'(x) &= \frac{d}{dx} \left[\frac{\sin(2x)}{x} \right] \\ \\ &= \frac{\left(\frac{d}{dx} \sin(2x) \right)x – \sin(2x) \left( \frac{d}{dx} x \right) }{x^2} \\ \\ &= \frac{\left(\cos(2x) \cdot \left(\frac{d}{dx} (2x) \right) \right) x – \sin(2x) (1) }{x^2} \\ \\ &= \frac{\left(\cos(2x) \cdot (2) \right) x – \sin(2x)}{x^2} \\ \\ &= \frac{2x\cos(2x) – \sin(2x)}{x^2} \quad \cmark \end{align*}

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We’ll solve this two ways, first with the Quotient Rule, and then with the Product Rule. Method 1: Quotient Rule \begin{align*} g'(x) &= \frac{d}{dx} \left[\frac{x^2}{\sqrt{1-x}}\right] \\ \\ &= \frac{\left(\frac{d}{dx}x^2 \right)\left(\sqrt{1-x}\right) – \left(x^2\right) \left[\frac{d}{dx}\sqrt{1-x}\right ]}{\left(\sqrt{1-x} \right)^2} \\ \\ &= \frac{(2x)\sqrt{1-x} – x^2\left[\frac{1}{2}\frac{1}{\sqrt{1-x}}\cdot \dfrac{d}{dx}(1-x)\right]}{1-x} \\ \\ &= \frac{(2x)\sqrt{1-x} – x^2\left[\frac{1}{2}\frac{1}{\sqrt{1-x}}\cdot (-1)\right]}{1-x} \\ \\ &= \frac{(2x)\sqrt{1-x} + \frac{x^2}{2\sqrt{1-x}}}{1-x} \\ \\ &= \dfrac{\dfrac{4x(1-x) +x^2}{2\sqrt{1-x}}}{1-x} \\ \\ &= \dfrac{4x -4x^2 + x^2}{2(1-x)\sqrt{1-x}} \\ \\ &= \frac{4x – 3x^2}{2(1-x)^{3/2}} \quad \cmark \end{align*} Method 2: Product Rule \begin{align*} g'(x) &= \frac{d}{dx}\left[x^2(1-x)^{-1/2}\right] \\ \\ &= \left[\frac{d}{dx}x^2\right](1-x)^{-1/2} + x^2 \frac{d}{dx}\left[(1-x)^{-1/2} \right] \\ \\ &= [2x](1-x)^{-1/2} + x^2 \left[\left(-\dfrac{1}{2}\right)(1-x)^{-3/2}\cdot\frac{d}{dx}[1-x]\right] \\ \\ &= [2x](1-x)^{-1/2} + x^2 \left[\left(-\dfrac{1}{2}\right)(1-x)^{-3/2}\cdot(-1)\right] \\ \\ &= \frac{2x}{(1-x)^{1/2}} + \dfrac{x^2}{2(1-x)^{3/2}} \\ \\ &= \frac{4x(1-x)}{2(1-x)^{3/2}} + \frac{x^2}{2(1-x)^{3/2}} \\ \\ &= \frac{4x – 4x^2 + x^2}{2(1-x)^{3/2}} \\ \\ &= \frac{4x – 3x^2}{2(1-x)^{3/2}} \quad \cmark \end{align*}

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The solution is easiest to understand if we rewrite the given derivative as $f'(u) = \sqrt{3u + 5},$ and then note that we have $u = x^2.$ Why can we write u instead of x for $f’$ here? Remember that the input variable is really a dummy variable, and we’re free to call it whatever we want. We think using a variable other than x in this first part makes things much clearer.

Because now we can invoke the Chain Rule, using $\dfrac{dy}{du} = f'(u) = \sqrt{3u + 5}:$ \begin{align*} \dfrac{dy}{dx} &= \dfrac{dy}{du} \cdot \dfrac{du}{dx} \\[8px] &= \sqrt{3u + 5} \cdot\dfrac{d}{dx}\left( x^2\right) \\[8px] &= \sqrt{3x^2 + 5} \cdot(2x) \\[8px] &= 2x \sqrt{3x^2 + 5} \quad \cmark \end{align*}

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A less formal approach, using “stuff”: We can view the function $f(x^2)$ as being $f(\text{stuff})$, where $\text{“stuff”} = x^2$. Then \begin{align*} \frac{dy}{dx} &= \frac{d}{dx} \left[f(\text{stuff})\right] \\ \\ &= f'(\text{stuff}) \cdot \left[\frac{d}{dx} \text{(stuff)} \right] \\ \\ &= f'(x^2) \cdot \left[\frac{d}{dx}(x^2) \right] \\ \\ &= \sqrt{3x^2 + 5} \cdot (2x) \quad \cmark \end{align*}

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