Question

log(secx+tanx) differencetion with respect to x​

Answer 1

Answer:

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Answer 2

Given :-

$log(sec \: x + tan \: x)$

To Find :-

differentiation With respect to x.

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$\huge{\underline{\underline{\purple{Solution→}}}}$

Let ,

$y = log(sec \: x + tan \: x)$

By Differentiating with respect to x -

We know that →

• $\frac{d}{dx} ( log(x) ) = \frac{1}{x}$

$= > \frac{dy}{dx} = \frac{1}{sec \: x + tan \: x} \times \frac{d}{dx} (sec \: x + tan \: x)$

Now -

• $\frac{d}{dx} (sec \: x) = sec \: x \: tan \: x$
• $\frac{d}{dx} (tan \: x) = {sec}^{2}x$

$= > \frac{dy}{dx} = \frac{1}{sec \: x + tan \: x} (sec \: x \: tan \: x + {sec}^{2} x) \\ \\ = > \frac{dy}{dx} = \frac{sec \: x(sec \: x + tan \: x)}{(sec \: x + tan \: x)}$

$\boxed{ = > \frac{dy}{dx} = sec \: x }$

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