Question

find the double derivative of y=secx/e^x​

Answer 1

AnswEr :

We have,

$\sf \: y = \dfrac{sec \: x}{e {}^{x} }$

For simplicity,let's rewrite the equation as :

$\implies \sf \: y = {e}^{ - x} sec \: x = uv$

The following expression is of the form uv,it's derivative would be of the form :

$\sf y' = u'v - v'u$

First Order Derivative would be :

$\longrightarrow \: \sf \: y' = \dfrac{d( {e}^{ - x} )}{dx} sec \: x + \dfrac{d(sec \: x)}{dx} e {}^{ - x} \\ \\ \longrightarrow \sf y' = - e {}^{ - x} sec \: x + {e}^{ - x} sec \: x.tan \: x \\ \\ \longrightarrow \sf \: y' = {e}^{ - x} sec \: x(tan \: x - 1)$

Similarly,

Second Order Derivative would be :

$\longrightarrow \sf \: y''= \dfrac{d({e}^{ - x})}{dx} sec \: x(tan \: x - 1) + \dfrac{d(tan \: x - 1)}{dx} e {}^{ - x} sec \: x \\ \\ \longrightarrow \sf \: y'' = e {}^{ - x} sec \: x(tan \: x - 1) {}^{2} + e {}^{ - x} sec {}^{3} x \\ \\ \longrightarrow \boxed{ \boxed{\sf \: y'' = e {}^{ - x} sec \: x \bigg((tan \: x - 1) {}^{2} + {sec}^{2} x \bigg)}}$

Answer 2

Answer:

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Explanation:

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