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If #f(x) =xe^x# and #g(x) = sinx-x#, what is #f'(g(x)) #?

Answer 1

Ezra Adams

#f'(g(x))=(cosx-1)e^(sinx-x)(sinx-x+1)#

#f(x)=xe^x# #g(x)=sinx-x#

#f(g(x))# means that we sub #sinx-x# into any #x# in #f(x)# #f(g(x))=(sinx-x)e^(sinx-x)#

#f'(g(x))=(sinx-x)times(cosx-1)e^(sinx-x)+e^(sinx-x)times (cosx-1)#

#f'(g(x))=(sinx-x)(cosx-1)e^(sinx-x)+(cosx-1)e^(sinx-x)#

#f'(g(x))=(cosx-1)e^(sinx-x)(sinx-x+1)#

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Answer 2

Charles Arocha

To find ( f'(g(x)) ), you first find ( f'(x) ), then substitute ( g(x) ) into ( f'(x) ).

Given ( f(x) = xe^x ) and ( g(x) = \sin x - x ):

Find ( f'(x) ): [ f'(x) = e^x + xe^x ]
Substitute ( g(x) ) into ( f'(x) ): [ f'(g(x)) = e^{\sin x - x} + (\sin x - x)e^{\sin x - x} ]

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Answer from HIX Tutor

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.