What is the derivative of $f (x) = e^{- x} cos (x^{2}) + e^{x} sin (x)$ $f(x) = e^-xcos(x^2)+e^xsin(x)$ ?

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What is the derivative of $f (x) = e^{- x} cos (x^{2}) + e^{x} sin (x)$ $f(x) = e^-xcos(x^2)+e^xsin(x)$ ?

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Answer 1 · 2016-07-09T13:01:55

$= - e^{- x} cos (x^{2}) - 2 x e^{- x} sin (x^{2}) + e^{x} sin (x) + e^{x} cos (x)$ $=-e^(-x)cos(x^2) - 2xe^(-x)sin(x^2) + e^xsin(x) + e^xcos(x)$

Explanation:

Product rule is our friend here.

$\frac{d}{d x} (e^{- x} cos (x^{2})) + \frac{d}{d x} (e^{x} sin (x))$ $d/(dx) (e^(-x)cos(x^2)) + d/(dx)(e^xsin(x))$

$= \frac{d}{d x} (e^{- x}) cos (x^{2}) + e^{- x} \frac{d}{d x} (cos (x^{2})) + \frac{d}{d x} (e^{x}) sin (x) + e^{x} \frac{d}{d x} (sin (x))$ $=d/(dx)(e^(-x))cos(x^2) + e^(-x)d/(dx)(cos(x^2)) + d/(dx)(e^x)sin(x) + e^xd/(dx)(sin(x))$

$= - e^{- x} cos (x^{2}) - 2 x e^{- x} sin (x^{2}) + e^{x} sin (x) + e^{x} cos (x)$ $=-e^(-x)cos(x^2) - 2xe^(-x)sin(x^2) + e^xsin(x) + e^xcos(x)$

NB: for $\frac{d}{d x} (cos (x^{2}))$ $d/(dx)(cos(x^2))$ I have used the chain rule because $x^{2}$ $x^2$ is also a function of x

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What is the derivative of $f (x) = e^{- x} cos (x^{2}) + e^{x} sin (x)$ $f(x) = e^-xcos(x^2)+e^xsin(x)$ ?

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Explanation:

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