The goal of this page is for you to practice using the Chain Rule on some beginning problems (all free, of course!), all with complete solutions. We’ll move to more complex ones on the next page, but you’ll probably find that you make some mistakes here, as have we all when first learning this stuff. If so, great! There’s no penalty, and most importantly, the only way to learn is through practice, so go ahead and do your early learning right here.
And even though we’re saying these are “beginning problems,” you’ll find some toward the bottom that are from past exams at some of the world’s best-known science and engineering universities.
We’ll start by seeing how the Chain Rule works with the Power Rule, Exponentials, Trig Functions, and then the Product and Quotient Rules.
Show/Hide Chain Rule Summary
The Chain Rule using prime notation: \begin{align*}\Big[ f\Big(g(x)\Big)\Big]’ &= 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*} In Leibniz notation: \[\dfrac{dy}{dt} = \dfrac{dy}{du} \cdot \dfrac{du}{dt} \]
And informally, the way you may quickly come to think about it: \[\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}\]
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Chain Rule and Power Rule
\begin{align*} \text{If} && f(x) &= (\text{stuff})^n, \\[8px]
\text{then} &&\dfrac{df}{dx} &= n(\text{that stuff})^{n-1} \cdot \dfrac{d}{dx}(\text{that stuff}) \end{align*} You’ll usually see this written as $$\dfrac{d}{dx}\left(u^n \right) = n u^{n-1} \cdot \dfrac{du}{dx}$$ The following problems illustrate and let you practice.
Chain Rule & Power Rule Problem #1
Given $f(x) = \left(3x^2 - 4x + 5\right)^8,$ $f'(x) =$
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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[hide solution]
Chain Rule & Power Rule Problem #2
Given $f(x) = \tan^3 x,$ $f'(x) =$
Hint: Recall $\tan^3 x = \big[\tan x\big]^3.$ Also recall that $\dfrac{d}{dx}\tan x = \sec^2 x.$
\begin{array}{lllll} \text{(A) }3\sec^4 x && \text{(B) }3\tan^2 x && \text{(C) }\tan^3 x \sec^2 x && \text{(D) }3\tan^2 x \cdot \sec^2 x && \text{(E) none of these} \end{array}
Nice.
Show/Hide Solution
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.”
(Open to develop those derivatives.)
Recall that $\dfrac{d}{du}\left(u^n\right) = nu^{n-1}.$ The rule also holds for fractional powers:
\begin{align*}
\dfrac{d}{du}\left(\sqrt{u} \right) &= \dfrac{d}{du}\left(u^{\frac{1}{2}} \right) \\[8px]
&= \frac{1}{2}u^{\left(\frac{1}{2} – 1 \right)} \\[8px]
&= \frac{1}{2}u^{-1/2} \\[8px]
&= \frac{1}{2}\frac{1}{\sqrt{u}}
\end{align*}
And let’s show $\left(x^2 + 1 \right)’ = 2x:$
\[ \begin{align*}
\dfrac{d}{dx}\left(x^2 + 1 \right) &= \dfrac{d}{dx}\left( x^2\right) + \cancelto{0}{\dfrac{d}{dx}(1)} \\[8px]
&= 2x^1 \\[8px]
&= 2x
\end{align*} \]
[collapse]
\[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}\]
\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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[hide solution]
Chain Rule & Power Rule Problem #4
Given $f(x) = \sqrt{\sin x}, \, f'(x) =$
\begin{array}{ll} \text{(A) } \dfrac{1}{2}\dfrac{1}{\sqrt{\sin x}} && \text{(B) } \dfrac{1}{2}\sqrt{\sin x}\cdot \cos x && \text{(C) } \dfrac{1}{2}\dfrac{1}{\sqrt{\sin x}}\cdot \cos x \end{array}
\begin{array}{ll} \text{(D) } -\dfrac{1}{2}\dfrac{1}{\sqrt{\sin x}}\cdot \cos x && \text{(E) none of these} \end{array}
Great
Show/Hide Solution
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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[hide solution]
Chain Rule and Exponentials
\begin{align*} \text{If} && f(x) &= e^{\text{(stuff)}}, \\[8px]
\text{then} &&\dfrac{df}{dx} &= e^{\text{(that stuff)}}\cdot \dfrac{d}{dx}(\text{that stuff}) \end{align*} You’ll usually see this written as $$\dfrac{d}{dx}e^u = e^u \cdot \dfrac{du}{dx}$$ The following problems illustrate and let you practice.
Chain Rule & Exponentials Problem #1
Given $f(x) = e^{\sin x}, \, f'(x) =$
\begin{array}{lllll} \text{(A) }e^{\cos x} \cos x && \text{(B) }e^{(\sin x -1)}e^{\cos x} && \text{(C) }e^{-\sin x} \cos x && \text{(D) }e^{\sin x} \cos x && \text{(E) none of these} \end{array}
You got it.
Show/Hide Solution
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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[hide solution]
Chain Rule and Trig Functions
\begin{align*} \text{If} && f(x) &= \sin\text{(stuff)}, \\[8px]
\text{then} &&\dfrac{df}{dx} &= \cos\text{(that stuff)}\cdot \dfrac{d}{dx}(\text{that stuff}) \end{align*} You’ll usually see this written as $$\dfrac{d}{dx}\sin u = \cos u \cdot \dfrac{du}{dx}$$
\begin{align*} \text{If} && f(x) &= \cos\text{(stuff)}, \\[8px]
\text{then} &&\dfrac{df}{dx} &= -\sin\text{(that stuff)}\cdot \dfrac{d}{dx}(\text{that stuff}) \end{align*} You’ll usually see this written as $$\dfrac{d}{dx}\cos u = -\sin u \cdot \dfrac{du}{dx}$$ \begin{align*} \text{If} && f(x) &= \tan\text{(stuff)}, \\[8px]
\text{then} &&\dfrac{df}{dx} &= \sec^2\text{(that stuff)}\cdot \dfrac{d}{dx}(\text{that stuff}) \end{align*} You’ll usually see this written as $$\dfrac{d}{dx}\tan u = \sec^2 u \cdot \dfrac{du}{dx}$$ The following problems illustrate and let you practice.
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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[hide solution]
Chain Rule and Product or Quotient Rule
The next few problems require using the Chain rule with the Product rule or with the Quotient rule.
Chain & Product, Quotient Rule Problem #1
This problem combines the Product Rule with the Chain Rule.
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:
$$\left[ \left(x^2 + 1 \right)^7\right]’ = 7\left(x^2 + 1 \right)^6 \cdot (2x) $$
Open for more detail
We can view $\left(x^2 + 1 \right)^7$ as $({\text{stuff}})^7$, where $\text{stuff} = x^2 + 1$. Then
\begin{align*}
f(x) &= (\text{stuff})^7; \quad \text{stuff} = x^2 + 1 \\[12px]
\text{Then}\phantom{f(x)= }\\
f'(x) &= 7(\text{stuff})^6 \cdot (x^2 + 1)’ \\[8px]
&= 7(x^2 + 1)^6 \cdot (2x)
\end{align*}
[collapse]
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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[hide solution]
Chain & Product, Quotient Rule Problem #2
This problem combines the Quotient rule with the Chain rule.
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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[hide solution]
Other Routine Chain Rule Problems
We’ll end this screen with some other typical Chain Rule problems you’re likely to encounter on an exam — including a few from actual university exams, which we hope will seem routine after the work you’ve done above. If they don’t seem routine yet, they will soon, as long as you keep practicing!
Chain Rule General Problem #1
Given that $f(2) = 1$, $f'(4) = 5$, $g(2) = 4$, and $g'(2) = 8$, find $\left[ f\big(g(2)\big)\right]'$.
\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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[hide solution]
Chain Rule General Problem #2
If $f(x)$ and $g(x)$ are represented in the graph, find $\Big[ f\Big(g(x)\Big) \Big]'$ at $x=0$.
\begin{array}{lllll} \text{(A) }0 && \text{(B) }-1 && \text{(C) }1 && \text{(D) }-\dfrac{3}{2}
&& \text{(E) }\dfrac{1}{2} \end{array}
Check: where does the graph of <em>f</em> have zero slope? Does your choice have $f'(x) = 0$ at those same values of <em>x</em>?
Show/Hide Solution
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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[hide solution]
Chain Rule General Problem #3
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 $\displaystyle{\lim_{h \to 0} \frac{f(g(2+h)) - f(g(2))}{h}}.$
Show/Hide Solution
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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[hide solution]
Chain Rule General Problem #4
[This problem appeared on an exam at a well-known science and engineering university.] Differentiate $f(x) = \dfrac{\sin(2x)}{x}.$
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Chain Rule General Problem #5
[This problem appeared on an exam at a well-known science and engineering university.] Differentiate $g(x) = \dfrac{x^2}{\sqrt{1-x}}.$
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This next problem is a little different, since you’re given $f'(x)$ and then asked to find the derivative of $f(x^2).$ You may encounter a similar problem in your homework or on an exam; this problem was taken, in fact, from an exam at a well-known university.
Chain Rule General Problem #6 (prior uni exam question)
If $y = f(x^2)$ and $f'(x) = \sqrt{3x + 5}$, show that $\dfrac{dy}{dx} = 2x\sqrt{3x^2 + 5}$.
Show/Hide Solution
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*} 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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On the next screen we’ll introduce problems that require using the Chain Rule more than once. It’s a small step from what we’ve done above. Since for the rest of the course you’re going to need to take such derivatives quickly and correctly, please proceed there to practice as soon as you can. Do you have questions about any of the problems on this screen, or other Chain Rule problems you’re working on? If you post on the Forum, we’ll do our best to assist!
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