Are you working to use the Chain Rule in Calculus? Students often initially find it confusing; let’s break it down and develop an easy can’t-fail approach.

Matheno Essentials: Chain Rule

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Chain Rule & 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 six problems illustrate.

Chain 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 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 Problem #3

Given $f(x) = (\cos x - \sin x)^{-2}, \, f'(x)=$

\begin{array}{ll} \text{(A) }-2(\cos x - \sin x)^{-1} \cdot (-\sin x - \cos x) && \text{(B) }-2(\cos x - \sin x)^{-3} && \end{array}

\begin{array}{ll} \text{(C) }-2(\cos x - \sin x)^{-3} \cdot (-\sin x - \cos x) && \text{(D) }-2(-\sin x - \cos x)^{-3} \cdot (\cos x - \sin x) \end{array}

\[\text{(E) none of these}\]

Chain Rule Problem #4

Given $f(x) = \left(x^5 + e^x\right)^{99}, \, f'(x) =$

\begin{array}{lll} \text{(A) }99\left(x^5 + e^x \right)^{98} \cdot (5x^4 + e^x) && \text{(B) }99\left(x^5 + e^x \right)^{98} && \text{(C) }99\left(x^5 + e^x \right)^{98} \cdot (5x^4 + e^{x-1}) \end{array}

\begin{array}{ll} \text{(D) }99\left(5x^4 + e^x \right)^{98} && \text{(E) none of these} \end{array}

Chain Rule Problem #5

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 Problem #6

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}

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Chain Rule & Exponentials

Chain Rule Problem #7

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 Problem #8

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 Problem #9

Given $f(x) = e^{\left(x^7 - 4x^3 + x \right)}, \, f'(x) =$

\begin{array}{lll} \text{(A) }e^{\left(x^7 - 4x^3 + x \right)} \cdot \left(7x^6 - 12x^2 \right) && \text{(B) }e^{\left(x^7 - 4x^3 + x \right)} \cdot \left(7x^6 - 12x^2 + 1 \right) && \text{(C) }e^{\left(x^7 - 4x^3 + x \right)} \end{array}

\begin{array}{ll} \text{(D) }e^{\left(7x^6 - 12x^2 + 1 \right)} && \text{(E) none of these} \end{array}

Chain Rule & 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 next two problems illustrate.

Chain Rule Problem #10

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 Problem #11

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 Rule or Quotient Rule

Chain Rule Problem #12

*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 Rule Problem #13

*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

Using the Chain rule multiple times

Chain Rule Problem #14

Differentiate $f(x) = \cos(\tan(3x)).$

Chain Rule Problem #15

Differentiate $f(x) = \sqrt{\sin (5x)}.$

Chain Rule Problem #16

Differentiate $f(x) = \left(1 + \sin^9(2x + 3) \right)^2.$

More Problems; Actual Exam Questions

Chain Rule Problem #17

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 Problem #18

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

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

#19: Show, using the Chain rule, that...

Show the following, as requested.**(a)** Use the Chain Rule and the Product Rule to develop the Quotient Rule. Start from $\dfrac{f(x)}{g(x)} = f(x) \Big( g(x)\Big)^{-1}$.**(b)** Prove that the derivative of an even function is an odd function. [Recall that for an even function, $f(-x) = f(x)$.]**(c)** Prove that the derivative of an odd function is an even function. [Recall that for an odd function, $f(-x) = -f(x)$.]

#20: Derivative of a^x

Show that $\dfrac{d}{dx}a^x = a^x \ln a$, where $a$ is a constant and $a > 0$. (For example, $\dfrac{d}{dx}2^x = 2^x \ln 2.$)

*Hint:* Start with $a^x = e^{\ln a^x}$, and use $\dfrac{d}{dx}e^u = e^u \cdot \dfrac{du}{dx}.$

#21: University exam question

If $y = f(x^2)$ and $f'(x) = \sqrt{3x + 5}$, show that $\dfrac{dy}{dx} = 2x\sqrt{3x^2 + 5}$.

#22: Derive some trig derivatives

Derive the following, as requested.**(a)** Find the derivative of $\sec \theta = \dfrac{1}{\cos \theta}$. *Hint:* Remember the Chain Rule.**(b)** Find the derivative of $\cot \theta = \dfrac{\cos \theta}{\sin \theta} = (\cos \theta)(\sin \theta)^{-1}$. Use the Product Rule.**(c)** Find the derivative of $\cot \theta = \dfrac{\cos \theta}{\sin \theta}$. Use the Quotient Rule.

#23: More university exam questions

Find the requested information.**(a)** [This problem appeared on an exam at a well-known science and engineering university.] Differentiate $f(x) = \dfrac{\sin(2x)}{x}.$**(b)** Let $y = \cos\left(\sqrt{1 + \sqrt{x}}\right)$. Find $y'$.

#24: Still more uni exam questions

Find the derivative of the following functions.**(a)** $h(x) = e^{\sin(-x)}$**(b)** [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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Jump down this page to using the Chain rule and: [Power rule] [Exponentials] [Trig Functions] [Product rule & Quotient rule] [Chain rule multiple times] [More problems and University exam problems]

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