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 January 27th, 2019, 06:57 PM #1 Member   Joined: Feb 2018 From: Iran Posts: 52 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: $\displaystyle f(x)=f((x-2)+2)=e^{x-2}g((x-2)^2+1)=e^{x-2}g(x^2-4x+5)$ 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 08:27 PM. January 27th, 2019, 08:42 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,965 Thanks: 2214 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 Tags chain, rule Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Marcos tavarez Calculus 4 December 30th, 2016 04:28 AM Ku5htr1m Calculus 2 November 30th, 2016 03:55 PM ungeheuer Calculus 1 July 30th, 2013 05:10 PM unwisetome3 Calculus 4 October 19th, 2012 01:21 PM Peter1107 Calculus 1 September 8th, 2011 10:25 AM

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