Find the derivative of $\frac{x + cos x}{tan x}$ $(x+cosx)/tanx$ ?

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Answer 1 · 2018-02-07T07:57:14

Derivative is $\frac{tan x - sin x tan x + x {sec}^{2} x + sec x}{{tan}^{2} x}$ $(tanx-sinxtanx+xsec^2x+secx)/tan^2x$

We use quotient rule, which says that derivative of $\frac{f (x)}{g (x)}$ $(f(x))/(g(x))$ is

$(g(x)f'(x)-f(x)g'(x))/[g(x)]^2$

Hence derivative of $(x+cosx)/tanx$ is

$(tanx(1-sinx)-(x+cosx)xx(-sec^2x))/tan^2x$

= $(tanx-sinxtanx+xsec^2x+secx)/tan^2x$

Find the derivative of (x+cosx)/tanxx+cosxtanx(x+cosx)/tanx?