Question #321de

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Answer 1 · 2017-10-16T01:52:39

$e^sinx(cos^2x)-e^sinx(sinx)$

We know that the first derivative is $e^sinx(cosx)$

To get the second derivative we apply the product rule:

Where $f'(g)+g'(f)$

Let's say that $f$ is our $e^sinx$ and $g$ is $cosx$

We take the derivative of $f$ and multiply it by $g$ and add the derivative of $g$ multiplied by $f$

$e^sinx(cosx)(cosx)+(-sinx)(e^sinx)$

Simplify:

$e^sinx(cos^2x)-e^sinx(sinx)$