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differentiate of e^{sin x} + (tan x )^x

$L e t y = e^{\sin x} + {(\tan x)}^{x} N o w t a k e u = e^{\sin x} a n d v = {(\tan x)}^{x} t h e r e f o r e y = u + v \frac{d y}{d x} = \frac{d u}{d x} + \frac{d v}{d x} ...... (1) u = e^{\sin x} t a k i n g \log o n b o t h s i d e s \log u = \sin x \log e d i f f e r e n t i a t e w . r . t x \frac{1}{u} \frac{d u}{d x} = \cos x \frac{d u}{d x} = u \cos x = e^{\sin x} \cos x N o w v = {(\tan x)}^{x} t a k i n g \log o n b o t h s i d e s \log v = x \log \tan x d i f f e r e n t i a t e w . r . t x \frac{1}{v} \frac{d v}{d x} = (\log \tan x + x \sec x \cos e c x) \frac{d v}{d x} = v (\log \tan x + x \sec x \cos e c x) = {(\tan x)}^{x} (\log \tan x + x \sec x \cos e c x) N o w p u t i n (1) \frac{d y}{d x} = e^{\sin x} \cos x + {(\tan x)}^{x} (\log \tan x + x \sec x \cos e c x)$

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