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

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To find ( f'(g(x)) ), we first need to find the derivative of ( f(x) ) and then substitute ( g(x) ) into the derivative of ( f(x) ).

Given ( f(x) = -e^{2x-1} ) and ( g(x) = 5\sin^2(x^2) ):

The derivative of ( f(x) ) with respect to ( x ) is ( f'(x) = -2e^{2x-1} ).

Substituting ( g(x) ) into ( f'(x) ) gives:

( f'(g(x)) = -2e^{2(g(x))-1} ).

Now we need to express ( g(x) ) within the ( f'(x) ) expression:

( g(x) = 5\sin^2(x^2) ).

So, ( g(x) = 5\sin(x^2)^2 ).

Now we substitute ( g(x) ) into ( f'(x) ):

( f'(g(x)) = -2e^{2(5\sin(x^2)^2)-1} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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