# How do you differentiate  f(x)= (7e^x+x)^2  using the chain rule.?

Dec 18, 2015

$f ' \left(x\right) = 2 \left(7 {e}^{x} + x\right) \left(7 {e}^{x} + 1\right)$

#### Explanation:

With the chain rule, if you have some composition of functions that looks like

$f \left(x\right) = g \left(h \left(x\right)\right)$,

then the derivative $f ' \left(x\right)$ is equal to

$f ' \left(x\right) = \frac{d}{\mathrm{dh}} g \left(h\right) \cdot \frac{d}{\mathrm{dx}} h \left(x\right)$.

Essentially, differentiate the outside function while treating the whole inside function as if it's a single variable, and multiply it by the derivative of the inside function.

To illustrate what I mean, just imagine $h = 7 {e}^{x} + x$ for a second.

Then we have

$f \left(x\right) = g \left(h \left(x\right)\right) = {h}^{2}$.

Right? $g \left(h\right)$ is just all of $h$ squared, and so is $f \left(x\right)$.

So, to find the derivative, we'll just apply the formula we have above.

$f ' \left(x\right) = \frac{d}{\mathrm{dh}} g \left(h\right) \cdot \frac{d}{\mathrm{dx}} h \left(x\right)$

The derivative of $g \left(h\right) = {h}^{2}$ with respect to $h$ is just $2 h$. (power rule)

And, the derivative of $h$ with respect to $x$ is $h ' = 7 {e}^{x} + 1$.

Let's plug all that in:

$f ' \left(x\right) = 2 h \cdot h '$

$\to f ' \left(x\right) = 2 \left(7 {e}^{x} + x\right) \left(7 {e}^{x} + 1\right)$