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 January 27th, 2019, 07:57 PM #1 Member   Joined: Feb 2018 From: Iran Posts: 30 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, 09:16 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs We are given: $\displaystyle f(x+2)=e^xg(x^2+1)$ And so: $\displaystyle f(x)=f((x-2)+2)=e^{x-2}g((x-2)^2+1)=e^{x-2}g(x^2-4x+5)$ Hence: $\displaystyle f'(x)=e^{x-2}g(x^2-4x+5)+2e^{x-2}g'(x^2-4x+5)(x-2)$ And so: $\displaystyle f'(2)=e^{2-2}g(2^2-4(2)+5)+2e^{2-2}g'(2^2-4(2)+5)(2-2)=g(1)=5$ Does that make sense? Thanks from topsquark Last edited by skipjack; January 27th, 2019 at 09:27 PM.
 January 27th, 2019, 09:42 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,298 Thanks: 1971 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 $x = 0$, $f\,'(2) = g(1) = 5$. Thanks from topsquark

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