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

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Apr 4, 2018

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Read below.

Explanation:

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.

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