If #f(x)= sqrt(x-2 # and #g(x) = e^(2x #, what is #f'(g(x)) #?

Question

If #f(x)=  sqrt(x-2  # and #g(x)  = e^(2x  #, what is #f'(g(x))  #?

Paisley Johnson · Accepted Answer

#1/(2\sqrt(e^(2x)-2))# #f'(g(x))# means that we find #f'(x)# and substitute #g(x)# for #x#. Now, in order to solve #f'(x)#, we need to use the chain rule since it is a composite function. We recall the chain rule: #(du)/dx=(du)/(dv)*(dv)/dx#, where #u# and #v# are functions of #x#. For this function, #\sqrt(x-2)#, we say that #u# is #\sqrt(x-2)# and #v# is #x-2#. In order to find #(du)/dx#, we first find #(du)/(dv)#, which is #(d(\sqrt(x-2)))/(d(x-2))=1/2*(x-2)^(-1/2)=1/(2\sqrt(x-2))#. Next, we find #(dv)/dx#, which is #(d(x-2))/dx=1#. Multiplying these gives us #f'(x)#, #1/(2\sqrt(x-2))#. The only thing now to do is to substitute #g(x)# in for #x#. We get the final answer: #1/(2\sqrt(e^(2x)-2))#.

Charles Anctil · Answer

undefined To find $ f'(g(x)) $, first compute $ g'(x) $, then substitute $ g(x) $ into $ f'(x) $.

Given:

$ f(x) = \sqrt{x - 2} $

$ g(x) = e^{2x} $

First, find $ g'(x) $:

$ g'(x) = \frac{d}{dx}(e^{2x}) = 2e^{2x} $

Now, substitute $ g(x) $ into $ f'(x) $:

$ f'(g(x)) = \frac{d}{dx}\left(\sqrt{g(x) - 2}\right) $

$ f'(g(x)) = \frac{d}{dx}\left(\sqrt{e^{2x} - 2}\right) $

Using the chain rule, $ \frac{d}{dx}(u^{1/2}) = \frac{1}{2}u^{-1/2}\frac{du}{dx} $, where $ u = e^{2x} - 2 $:

$ f'(g(x)) = \frac{1}{2}\left(e^{2x} - 2\right)^{-1/2} \cdot \frac{d}{dx}(e^{2x} - 2) $

Now find $ \frac{d}{dx}(e^{2x} - 2) $:

$ \frac{d}{dx}(e^{2x} - 2) = 2e^{2x} $

Substitute back:

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

$ f'(g(x)) = \frac{e^{2x}}{\sqrt{e^{2x} - 2}} $