Find the derivative of frac{x+cos x}{tan x}. | vedclass

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(A) Let $f(x) = \frac{x+\cos x}{	an x}$.Using the quotient rule,$\frac{d}{dx} \left( \frac{u}{v} ight) = \frac{u'v - uv'}{v^2}$,where $u = x+\cos x

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$\frac{(1-\sin x) 	an x-(x+\cos x) \sec ^{2} x}{	an^{2} x}$

Vedclass · Answer

(A) Let $f(x) = \frac{x+\cos x}{	an x}$.Using the quotient rule,$\frac{d}{dx} \left( \frac{u}{v} ight) = \frac{u'v - uv'}{v^2}$,where $u = x+\cos x$ and $v = 	an x$.$u' = \frac{d}{dx}(x+\cos x) = 1 - \sin x$.$v' = \frac{d}{dx}(	an x) = \sec^2 x$.Substituting these into the formula:$f'(x) = \frac{(1-\sin x) 	an x - (x+\cos x) \sec^2 x}{(	an x)^2}$.Thus,the derivative is $\frac{(1-\sin x) 	an x - (x+\cos x) \sec^2 x}{	an^2 x}$.

Find the derivative of $\frac{x+\cos x}{\tan x}$.

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