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

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

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Answer 1 · 2018-03-02T17:13:16

$\frac{d y}{d x} = 2 x e^{x^{2} cos x} (x sin x - 2 cos x)$ $dy/dx=2xe^(x^2cosx)(xsinx-2cosx)$

Explanation:

$f (x) = - 2 e^{x^{2} cos x}$ $f(x)=-2e^(x^2cosx)$
Let
$y = f (x)$ $y=f(x)$
Then
$y = - 2 e^{x^{2} cos x}$ $y=-2e^(x^2cosx)$
Let
$u = e^{x^{2} cos x}$ $u=e^(x^2cosx)$
Taking logarithms
$ln u = x^{2} cos x$ $lnu=x^2cosx$
Differentiating wrt x
$\frac{1}{u} \frac{d u}{d x} = x^{2} (- sin x) + cos x (2 x)$ $1/u(du)/dx=x^2(-sinx)+cosx(2x)$
$\frac{1}{u} \frac{d u}{d x} = - x^{2} sin x + 2 x cos x$ $1/u(du)/dx=-x^2sinx+2xcosx$
$\frac{d u}{d x} = u (- x^{2} sin x + 2 x cos x)$ $(du)/dx=u(-x^2sinx+2xcosx)$
$y = - 2 e^{x^{2} cos x}$ $y=-2e^(x^2cosx)$
$y = - 2 u$ $y=-2u$
Differentiating
$\frac{d y}{d x} = - 2 \frac{d u}{d x}$ $dy/dx=-2(du)/dx$
$\frac{d u}{d x} = u (- x^{2} sin x + 2 x cos x)$ $(du)/dx=u(-x^2sinx+2xcosx)$
$\frac{d y}{d x} = - 2 (u (- x^{2} sin x + 2 x cos x))$ $dy/dx=-2(u(-x^2sinx+2xcosx))$
$\frac{d y}{d x} = 2 u (x^{2} sin x - 2 x cos x)$ $dy/dx=2u(x^2sinx-2xcosx)$
$u = e^{x^{2} cos x}$ $u=e^(x^2cosx)$
$\frac{d y}{d x} = 2 e^{x^{2} cos x} (x^{2} sin x - 2 x cos x)$ $dy/dx=2e^(x^2cosx)(x^2sinx-2xcosx)$
$\frac{d y}{d x} = 2 x e^{x^{2} cos x} (x sin x - 2 cos x)$ $dy/dx=2xe^(x^2cosx)(xsinx-2cosx)$

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

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Explanation:

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How do you differentiate f(x)=-2e^(x^2cosx f(x)=−2ex2cosxf(x)=-2e^(x^2cosx using the chain rule?

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