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

Question

Emma Aldridge · Accepted Answer

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(x) = g(x) * h(x)# then,

#f'(x) = g'(x)*h(x) + h'(x)*g(x)#

Lets get started!

In this example, lets just change #g(x)# to #f(x)# so it fits with the general formula. With this, your g(x) and h(x) are as follows: #g(x) = e^x# #h(x) = sqrt(1-e^2x)# which is equivalent to #(1-e^2x)^(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'(x) = e^x -> # Note that the derivative of #e^x# is always #e^x# #h'(x) = 1/2(1-e^2x)^(-1/2)*(-2e^(2x)) -> # Computed using chain rule! Now, sub everything in: #f'(x) = g'(x)*h(x) + h'(x)*g(x)# #f'(x) = (e^x)((1-e^2x)^(1/2)) +(1/2(1-e^2x)^(-1/2)*(-2e^(2x)))(e^x)# 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! :)

Christopher Allers · Answer

undefined To differentiate $ g(x) = e^x \sqrt{1 - e^{2x}} $ using the product rule, let $ u(x) = e^x $ and $ v(x) = \sqrt{1 - e^{2x}} $. Then apply the product rule:

$$ g'(x) = u'(x)v(x) + u(x)v'(x) $$

Where $ u'(x) = e^x $ and $ v'(x) = \frac{-e^{2x}}{2\sqrt{1 - e^{2x}}} $.

Plug these into the formula:

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