If y- log x-tan x- then dydx-

The correct option is A x y=log( x+tan x) dydx=1 x+tan x(secx+tan x)' #160; #160; #160; #160; =1 x+tan x( x tan x+^2x) #160; ...

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Integration Using Substitution

If y= log ...

Question

If $y$ = $log (sec x + tan x)$ , then $\frac{d y}{d x} =$

sec x

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\frac{1}{sec x + tan x}

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log (cos x + {sec}^{2} x)

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none of these

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Solution

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Similar questions

y = l o g (s e c x + tan x),

\frac{d y}{d x} =

If $y = [\frac{x^{2} + 1}{x + 1}], t h e n \frac{d y}{d x}$ =?

y = l o g \sqrt{\frac{1 + tan x}{1 - tan x}}

\frac{d y}{d x} = sec 2 x

y = {tan}^{- 1} (sec x + tan x)

\frac{d y}{d x} =

$I f y = 3^{l o g_{9}^{(1 + t a n^{2} z)}} t h e n \frac{d y}{d x} =$

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