If #f(x) =tan^2x # and #g(x) = sqrt(5x-1 #, what is #f'(g(x)) #?

1 Answer
Morgan
Aug 2, 2017

#f'(g(x))=(5sec^2(sqrt(5x-1))tan(sqrt(5x-1)))/(sqrt(5x-1))#

Explanation:

To find #f'(g(x))#, we can start by finding #f(g(x))# and then take the derivative. We substitute #sqrt(5x-1)# into #f(x)#, so we have:

#f(g(x))=f(sqrt(5x-1))=tan^2(sqrt(5x-1))#

Now we take the derivative using the chain rule.

#2tan(sqrt(5x-1))*sec^2(sqrt(5x-1))*1/2(5x-1)^(-1/2)*5#

#=>(5sec^2(sqrt(5x-1))tan(sqrt(5x-1)))/(sqrt(5x-1))#

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