Question

find d/dx of log(secx+tanx)

Answer 1

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Answer 2

Answer:

The d/dx of log(secx+tanx) is sec x.

Step-by-step explanation:

The given expression is

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

we have to find the d/dx of given expression.

$\frac{d}{dx}log(\sec x+\tan x)=\frac{1}{(\sec x+\tan x)}\times \frac{d}{dx}(\sec x+\tan x)$          $[\because \frac{d}{dx}log x=\frac{1}{x}]$        

$\frac{d}{dx}log(\sec x+\tan x)=\frac{1}{(\sec x+\tan x)}\times (\sec x\tan x+\sec^2x)$          $[\because \frac{d}{dx}\sec x=\sec x\tan x,\frac{d}{dx}\tan x=\sec^2x]$  

$\frac{d}{dx}log(\sec x+\tan x)=\frac{\sec x(\sec x+\tan x)}{(\sec x+\tan x)}$

Cancel out the common factor.

$\frac{d}{dx}log(\sec x+\tan x)=\sec x$

Therefore the d/dx of log(secx+tanx) is sec x.