# Math Insight

### Simple examples of using the chain rule

The chain rule is a formula to calculate the derivative of a composition of functions. Once you have a grasp of the basic idea behind the chain rule, the next step is to try your hand at some examples.

#### Example 1

Let $f(x)=6x + 3$ and $g(x)=-2x+5$. Use the chain rule to calculate $h'(x)$, where $h(x)=f(g(x))$.

Solution: The derivatives of $f$ and $g$ are \begin{align*} f'(x)&=6\\ g'(x)&=-2. \end{align*} According to the chain rule, \begin{align*} h'(x) &= f'(g(x)) g'(x)\\ &= f'(-2x +5) (-2)\\ &= 6 (-2)=-12. \end{align*}

Since the functions were linear, this example was trivial. Even though we had to evaluate $f'$ at $g(x)=-2x+5$, that didn't make a difference since $f'=6$ not matter what its input is. Moveover, in this case, if we calculate $h(x)$, \begin{align*} h(x) &= f(g(x))\\ &= f(-2x+5)\\ &= 6(-2x+5)+3\\ &= -12x+30+3 = -12x + 33, \end{align*} then we can quite easily calculate its derivative directly to obtain that $h'(x)=-12$.

#### Example 2

Let $f(x)=e^x$ and $g(x)=4x$. Use the chain rule to calculate $h'(x)$, where $h(x)=f(g(x))$.

Solution: The derivative of the exponential function with base $e$ is just the function itself, so $f'(x)=e^x$. The derivative of $g$ is $g'(x)=4$. According to the chain rule, \begin{align*} h'(x) &= f'(g(x)) g'(x)\\ &= f'(4x) \cdot 4\\ &= 4e^{4x}. \end{align*}

In this example, it was important that we evaluated the derivative of $f$ at $4x$. The derivative of $h(x)=f(g(x))=e^{4x}$ is not equal to $4e^{x}$. The only correct answer is $h'(x)=4e^{4x}$. This result is a special case of equation (5) from the derivative of exponential function page with $k=4$.

#### Example 3

Let $f(x)=e^x$ and $g(x)=3x^2+2$. Use the chain rule to calculate $h'(x)$, where $h(x)=f(g(x))$.

Solution: The derivatives of $f$ and $g$ are \begin{align*} f'(x)&=e^x\\ g'(x)&=6x. \end{align*} According to the chain rule, \begin{align*} h'(x) &= f'(g(x)) g'(x)\\ &= f'(3x^2+2) (6x)\\ &= 6x e^{3x^2+2} \end{align*}

#### Example 4

Calculate the derivative of $h(x)=e^{3x^2+2}$.

Solution: This problem is a chain rule problem in disguise. We don't know how to calculate the derivaitve of $e^{3x^2+2}$ directly. However, we do know how to calculate the derivative of $e^x$ and $3x^2+2$. If we let $f(x)=e^x$ and $g(x)=3x^2+2$, then $h(x)=f(g(x))$. This problem is the same as the previous example in disguise. The derivative is $h'(x)=6xe^{3x^2+2}$.

#### Example 5

Calculate the derivative of $f(x)=6e^{x^4+x^2}$

Solution: Again, we must use the chain rule. The function $f$ is the composition of the function $g(z)=6e^z$ and $h(x)=x^4+x^2$. (It's OK if we use different notation for the functions or the inputs of the functions.) With these definitions of $g$ and $h$, our function $f$ is $f(x)=g(h(x))$.

The derivatives of the component functions are \begin{align*} g'(z) &= 6e^z\\ h'(x)&=4x^3+2x \end{align*} so that the derivative of $f$ is \begin{align*} f'(x) &= g'(h(x))h'(x) \\ &= g'(x^4+x^2) (4x^3+2x)\\ &= 6e^{x^4+x^2}(4x^3+2x)\\ &= (24 x^3 + 12 x) e^{x^4+x^2} \end{align*}

#### Example 6

Calculate the derivative of $g(x)=\ln (x^2+1)$.

Solution: To use the chain rule for this problem, we need to use the fact that the derivative of $\ln(z)$ is $1/z$. Then, by the chain rule, the derivative of $g$ is \begin{align*} g'(x) &= \diff{}{x}\ln(x^2+1) \\ &= \frac{1}{x^2+1} (2x)\\ &= \frac{2x}{x^2+1}. \end{align*}

In this example, we didn't bother specifying the component functions by denoting them with a letter but used the expression $\diff{}{x}(\text{stuff})$ to indicate the derivative of “stuff” with respect to $x$. We could have, for example, let $p(z)=\ln(z)$ and $q(x)=x^2+1$ so that $p'(z)=1/z$ and $q'(x)=2x$. Then, we could have written $g'(x)$ as $p'(q(x))q'(x) = \frac{1}{x^2+1} (2x)$. Typically, when using the chain rule, we won't bother with the extra steps of defining the component functions.

#### Example 7

Calculate the derivative of $\ln (1+e^y)$ with respect to $y$.

Solution: Since we know the derivative of $\ln(z)$ is $1/z$ and the derivative of $1+e^y$ is $e^y$, we can use the chain rule to calculate that \begin{align*} \diff{}{y} \ln(1+e^y) &= \frac{1}{1+e^y}(e^y)\\ &= \frac{e^y}{1+e^y}. \end{align*}

#### Example 8

Calculate $\displaystyle\diff{}{x}(2x^2+x)^3$.

Solution: \begin{align*} \diff{}{x}(2x^2+x)^3 &=3(2x^2+x)^2(4x+1) \end{align*}

For additional examples, see the chain rule page from the Calculus Refresher.