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January 27th, 2019, 06:57 PM   #1
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Unhappy 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
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January 27th, 2019, 08:16 PM   #2
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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?
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Last edited by skipjack; January 27th, 2019 at 08:27 PM.
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January 27th, 2019, 08:42 PM   #3
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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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