If $f(x)= cot2 x$ and $g(x) = e^(1 - 4x )$ , how do you differentiate $f(g(x))$ using the chain rule?

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Answer 1 · 2016-10-13T19:15:20

$(8e^(1-4x))/sin^2(2e^(1-4x))$ or $8e^(1-4x)csc^2(2e(1-4x))$

$f(g(x))=cot2e^(1-4x)$

Let $g(x)=u$

$f'(u)=d/(du)cot2u=d/(du)(cos2u)/(sin2u)=(-2sin(2u)sin(2u)-2cos(2u)cos(2u))/sin^2(2u)$
$=(-2sin^2(2u)-2cos^2(2u))/sin^2(2u)$
$=-2/sin^2(2u)$

$g'(x)=-4e^(1-4x)$

Using chain rule: $f'(g(x))=f'(u)*g'(x)$

$=-2/sin^2(2u)*-4e^(1-4x)$

$=-2/sin^2(2e^(1-4x))*-4e^(1-4x)$

$=(8e^(1-4x))/sin^2(2e^(1-4x))$ or $8e^(1-4x)csc^2(2e(1-4x))$

If f(x)= cot2 x f(x)= cot2 x and g(x) = e^(1 - 4x ) g(x) = e^(1 - 4x ) , how do you differentiate f(g(x)) f(g(x)) using the chain rule?