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

1 Answer
Mar 27, 2017

Answer:

#1/(2\sqrt(e^(2x)-2))#

Explanation:

#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))#.