Question

$if \: y = {x}^{ \tan(x) } + { \sin(?) }^{cos(x) } find \: dy \div dx$
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Answer 1

Step-by-step explanation:

We have,

$y = {x}^{ \tan(x) } + \sin(x) ^{ \cos(x) }$

$\implies \frac{dy}{dx} = {x}^{ \tan(x) } [ \frac{d}{dx} ( \tan(x) ln(x)) ] + { \sin(x) }^{ \cos(x) } [ \frac{d}{dx}( \cos(x) ln( \sin(x) ) )]$

$\implies \frac{dy}{dx} = {x}^{ \tan(x) } [ \sec^{2} (x) ln(x) + \frac{ \tan(x) }{x} ] + { \sin(x) }^{ \cos(x) } [ - \sin(x) ln( \sin(x) ) + \cos(x) \tan(x) ]$

Answer 2

Step-by-step explanation:

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