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If #f(x) =-e^(-2x-7) # and #g(x) = 3/x #, what is #f'(g(x)) #?

Answer 1

Christopher Agresta

#f'(g(x))=2e^(-6/x-7)#

First thing is to calculate the derivate of #f(x)#, #f'(x)=2e^(-2x-7)#

and now we substitute #g(x)# in every #x# of the #f'(x)#, #f'(g(x))=2e^(-2·3/x-7)=2e^(-6/x-7)#

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Answer 2

Samantha Adkison

To find ( f'(g(x)) ), we first need to find the composition of ( f ) and ( g ), denoted as ( f(g(x)) ), and then differentiate it with respect to ( x ).

[ f(g(x)) = f\left(\frac{3}{x}\right) = -e^{-2\left(\frac{3}{x}\right)-7} ]

Now, differentiate ( f(g(x)) ) with respect to ( x ) using the chain rule:

[ \frac{d}{dx}\left(-e^{-2\left(\frac{3}{x}\right)-7}\right) = -\frac{d}{dx}\left(e^{-2\left(\frac{3}{x}\right)-7}\right) = -\left(-2\cdot\frac{d}{dx}\left(\frac{3}{x}\right)\right)\cdot e^{-2\left(\frac{3}{x}\right)-7} ]

Using the chain rule again:

[ -\left(-2\cdot\frac{d}{dx}\left(\frac{3}{x}\right)\right)\cdot e^{-2\left(\frac{3}{x}\right)-7} = 2\cdot\frac{3}{x^2}\cdot e^{-2\left(\frac{3}{x}\right)-7} ]

Simplify:

[ f'(g(x)) = \frac{6}{x^2} \cdot e^{-2\left(\frac{3}{x}\right)-7} ]

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Answer from HIX Tutor

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