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January 27th, 2019, 06:57 PM  #1 
Member Joined: Feb 2018 From: Iran Posts: 50 Thanks: 3  Chain rule
If, f(x+2)=e^(x)*g(x^(2)+1) and g(1)=5 are given find f'(2)? Can you explain this step by step cause I have problem understanding chain rule 
January 27th, 2019, 08:16 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 
We are given: $\displaystyle f(x+2)=e^xg(x^2+1)$ And so: $\displaystyle f(x)=f((x2)+2)=e^{x2}g((x2)^2+1)=e^{x2}g(x^24x+5)$ Hence: $\displaystyle f'(x)=e^{x2}g(x^24x+5)+2e^{x2}g'(x^24x+5)(x2)$ And so: $\displaystyle f'(2)=e^{22}g(2^24(2)+5)+2e^{22}g'(2^24(2)+5)(22)=g(1)=5$ Does that make sense? Last edited by skipjack; January 27th, 2019 at 08:27 PM. 
January 27th, 2019, 08:42 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
It's necessary to assume that $g$ is differentiable at $x = 0$. Differentiating $f(x + 2) = e^xg(x^2 + 1)$ gives $f\,'(x + 2) = e^x(2x)g'(x^2+1) + e^xg(x^2 + 1)$, so by putting $x = 0$, $f\,'(2) = g(1) = 5$. 

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