How do you differentiate $f (x) = 3 x {(e^{{(x - 9)}^{2}})}^{3}$ $f(x) = 3x(e^((x-9)^2))^3$ using the chain rule?

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Answer 1 · 2017-06-01T04:42:00

$e^{{(x - 9)}^{6}} (18 x {(x - 9)}^{5} + 3)$ $e^((x-9)^6)(18x(x-9)^5+3)$

Step 1. Simplify the exponent

$f (x) = 3 x e^{{(x - 9)}^{6}}$ $f(x)=3x e^((x-9)^6)$

Step 2. Using the Product Rule directly uses the chain rule.

$f'(x)=3x d/dx(e^((x-9)^6))+e^((x-9)^6) d/dx(3x)$

$=3x e^((x-9)^6) d/dx((x-9)^6)+3e^((x-9)^6)$

$=3x e^((x-9)^6)(6(x-9)^5)+3e^((x-9)^6)$

$=18x(x-9)^5 e^((x-9)^6)+3e^((x-9)^6)$

Step 3. Factor out the common $e^((x-9)^6)$ term

$=e^((x-9)^6)(18x(x-9)^5+3)$

How do you differentiate f(x) = 3x(e^((x-9)^2))^3f(x)=3x(e(x−9)2)3f(x) = 3x(e^((x-9)^2))^3 using the chain rule?