If #f(x) =sinx # and #g(x) = (x+3)^3 #, what is #f'(g(x)) #?

Question

If #f(x) =sinx # and #g(x) = (x+3)^3  #, what is #f'(g(x)) #?

Aurora Alvarado · Accepted Answer

Read below.

Hmm... Let's let #g(x)=y# We now apply this to #f(x)#. #=>f(y)=sin(y)# Substitute #y# with #(x+3)^3# #=>f(g(x))=sin((x+3)^3)# If you meant to find just #f'(g(x))#... Just find the derivative of #sin(y)# Since derivative of #sin(x)# is #cos(x)#... #=>f'(g(x))=cos((x+3)^3)# If you meant to ask for #d/dx[f(g(x))],# you use the chain rule: #d/dx[f(g(x))]=f'(g(x))*g'(x)# We already know that #f'(g(x))=cos((x+3)^3)# #=>d/dx[f(g(x))]=cos((x+3)^3)*d/dx[(x+3)^3]# We use the power rule: #d/dx[x^n]=nx^(n-1)# if #n# is a constant. #=>d/dx[f(g(x))]=cos((x+3)^3)*3(x+3)^(3-1)*d/dx[x^1+3x^0]# Note here that the chain rule is applied again. #=>d/dx[f(g(x))]=cos((x+3)^3)*3(x+3)^2*1*x^(1-1)+3*0*x^(0-1)# #=>d/dx[f(g(x))]=cos((x+3)^3)*3(x+3)^2*1*1+0# #=>d/dx[f(g(x))]=cos((x+3)^3)*3(x+3)^2# That is the entire derivative calculated.

Emma Aldridge · Answer

undefined To find $ f'(g(x)) $, we first need to find $ g'(x) $ and then substitute it into $ f'(x) $ where $ x = g(x) $.

Given $ g(x) = (x+3)^3 $, differentiate it to find $ g'(x) $.

$ g'(x) = 3(x+3)^2 $

Now, since $ f(x) = \sin(x) $, $ f'(x) = \cos(x) $.

Substitute $ g(x) $ into $ f'(x) $:

$ f'(g(x)) = \cos(g(x)) = \cos((x+3)^3) $