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:

\[\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}\]

\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 $f(x) = \left(3x^2 - 4x + 5\right)^8,$ $f'(x) =$

\begin{array}{lll} \text{(A) }8\left(3x^2 - 4x + 5\right)^7 && \text{(B) }8\left(3x^2 - 4x + 5\right)^7 \cdot (6x -4) && \text{(C) }8(6x - 4)^7 \end{array}

\begin{array}{ll} \ \text{(D) }\left(3x^2 - 4x + 5\right)^8 && \text{(E) none of these} \end{array}

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}

Chain Rule & Power Rule Problem #3

Given $f(x) = \sqrt{x^2+1}, \, f'(x) =$

\begin{array}{lllll} \text{(A) }\dfrac{1}{2}\dfrac{1}{\sqrt{{x^2 + 1}}}\cdot ( 2x ) && \text{(B) }\dfrac{1}{2}\dfrac{1}{\sqrt{{x^2 + 1}}} && \text{(C) } \dfrac{1}{2}\dfrac{1}{\sqrt{{2x}}} && \text{(D) }\dfrac{1}{2}\sqrt{x^2 + 1} && \text{(E) none of these} \end{array}

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}

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}

Chain Rule & Exponentials Problem #2

Given $f(x) = e^{x^2}, \, f'(x) =$

\begin{array}{lllll} \text{(A) }e^{x^2}e^{2x} && \text{(B) }e^{x^2} \cdot 2x && \text{(C) }e^{x^2} && \text{(D) }e^{2x} && \text{(E) none of these} \end{array}

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.

Chain Rule & Trig Problem #1

Given $f(x) = \sin(2x), \, f'(x) =$

\begin{array}{lllll} \text{(A) }\cos(2x) \cdot(2) && \text{(B) }\sin(2x) \cdot (2) && \text{(C) }-\cos(2x)\cdot (2) && \text{(D) }\cos(2) && \text{(E) none of these} \end{array}

Chain Rule & Trig Problem #2

Given $f(x) = \tan\left(e^x\right), \, f'(x) =$ \begin{array}{lll} \text{(A) }\sec\left(e^x\right)\tan\left(e^x\right)\cdot e^x && \text{(B) }\sec\left(e^x\right)\cdot e^x && \text{(C) }\sec^2\left(e^x\right) \end{array} \begin{array}{ll} \text{(D) }\sec^2(e^x) \cdot e^x && \text{(E) none of these} \end{array}

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. *

Given $f(x) = \left(x^2 + 1 \right)^7 (3x - 7)^4, \, f'(x) =$

(A) $28\left(x^2 + 1 \right)^6 (3x - 7)^3$

(B) $28\left(x^2 + 1 \right)^6 (3x - 7)^3 \cdot (2x) \cdot (3)$

(C) $ \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]$

(D) $7\left(x^2 + 1 \right)^6 (3x - 7)^4 + 4\left(x^2 + 1 \right)^7 (3x - 7)^3$

(E) none of these

Chain & Product, Quotient Rule Problem #2

*This problem combines the Quotient rule with the Chain rule.*

Given $h(x) = \dfrac{e^{2x}}{1-x^2}, \, f'(x) =$

(A) $\dfrac{\left( e^{2x}\cdot 2\right)\left(1-x^2 \right) - \left(e^{2x} \right)\left(-2x \right)}{\left(1-x^2 \right)^2(-2x)}$

(B) $\dfrac{\left( e^{2x}\cdot 2\right)\left(1-x^2 \right) - \left(e^{2x} \right)\left(-2x \right)}{\left(1-x^2 \right)^2}$

(C) $\dfrac{\left( e^{2x}\cdot 2\right)\left(1-x^2 \right) - 2\left(e^{2x} \right)}{\left(1-x^2 \right)^2}$

(D) $\dfrac{\left( e^{2x}\cdot 2\right)\left(1-x^2 \right) + \left(e^{2x} \right)\left(-2x \right)}{\left(1-x^2 \right)^2}$

(E) none of these

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{array}{lllll} \text{(A) }5 && \text{(B) }10 && \text{(C) }40 && \text{(D) }8 && \text{(E) }32 \end{array}

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}

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 $\displaystyle{\lim_{h \to 0} \frac{f(g(2+h)) - f(g(2))}{h}}.$

Chain Rule General Problem #4

[This problem appeared on an exam at a well-known science and engineering university.] Differentiate $f(x) = \dfrac{\sin(2x)}{x}.$

Chain Rule General Problem #5

[This problem appeared on an exam at a well-known science and engineering university.] Differentiate $g(x) = \dfrac{x^2}{\sqrt{1-x}}.$

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 $\dfrac{dy}{dx} = 2x\sqrt{3x^2 + 5}$.

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!