## 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.

\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.

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

\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.

\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.

*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 $\sin^9(\cdots) = \big[\sin(\cdots) \big]^9.$

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