## Chain Rule: Problems and Solutions

### Chain Rule: Problems and Solutions

Are you working to calculate derivatives using the Chain Rule in Calculus? Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself.

Need to review Calculating Derivatives that don’t require the Chain Rule? That material is here.

Want to skip the Summary? Jump down to problems and their solutions.

You can always access our Handy Table of Derivatives and Differentiation Rules via the Key Formulas menu item at the top of every page.

\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 five problems illustrate.

Differentiate $f(x) = \left(3x^2 – 4x + 5\right)^8.$

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

*Hint:*Recall $\tan^3 x = \big[\tan x\big]^3.$ Also recall that $\dfrac{d}{dx}\tan x = \sec^2 x.$

Differentiate $f(x) = (\cos x – \sin x)^{-2}.$

Differentiate $f(x) = \left(x^5 + e^x\right)^{99}.$

Differentiate $f(x) = \sqrt{x^2+1}.$

\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 next two problems illustrate.

Differentiate $f(x) = e^{\sin x}.$

Differentiate $f(x) = e^{\left(x^7 – 4x^3 + x \right)}.$

\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.

Differentiate $f(x) = \sin(2x).$

Differentiate $f(x) = \tan(e^x).$

The problems below combine the Product rule and the Chain rule, or require using the Chain rule multiple times.

*This problem combines the Product Rule with the Chain Rule.*

Differentiate $f(x) = \left(x^2 + 1 \right)^7 (3x – 7)^4.$

*This problem requires using the Chain Rule twice.*

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

*This problem requires using the Chain Rule three times.*

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

*Hint:*Recall that $\sin^9(\cdots) = \big[\sin(\cdots) \big]^9.$

Need to use the derivative to find the equation of a tangent line (or the equation of a normal line)? We have a separate page on that topic here.

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Good ques but make ques a bit tough

Glad you thought these questions are good, and sounds like you’d like some more challenging ones to try for yourself. We’ll aim to add some!

Thanks very much for the comment! ðŸ™‚