# How do you differentiate g(x) = e^xsqrt(1-e^(2x)) using the product rule?

Hey there! To differentiate a function using the product rule, keep note of the general formula for the derivative of a product whereby if:

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

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

Lets get started!

#### Explanation:

In this example, lets just change $g \left(x\right)$ to $f \left(x\right)$ so it fits with the general formula. With this, your g(x) and h(x) are as follows:

$g \left(x\right) = {e}^{x}$

$h \left(x\right) = \sqrt{1 - {e}^{2} x}$ which is equivalent to ${\left(1 - {e}^{2} x\right)}^{\frac{1}{2}}$

Now, if you follow the derivative general formula, it reads "derivative of the 1st, times the 2nd - plus derivative of the 2nd times the 1st. Lets get those derivatives separately:

$g ' \left(x\right) = {e}^{x} \to$ Note that the derivative of ${e}^{x}$ is always ${e}^{x}$

$h ' \left(x\right) = \frac{1}{2} {\left(1 - {e}^{2} x\right)}^{- \frac{1}{2}} \cdot \left(- 2 {e}^{2 x}\right) \to$ Computed using chain rule!

Now, sub everything in:

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

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

And that's it! One suggestion I do have; if you can do these "inner" derivative in your head and as you go(i.e. the chain rule we had to do), this will allow you to complete the question much faster. I only did the derivatives separately for demonstrative purposes.

Hopefully this helped and was clear for you! If you have any questions, please let me know! :)