Question

Differentiate xsin x, x > 0 w.r.t. x.​

Answer 1

Answer:

sin x + x cosx

Step-by-step explanation:

d/dx xsinx

using product rule

= d/dx x*sin x + x * d/dx sin x

= 1*sin x + x cos x

= sin x + x cosx

Answer 2

$\sf \huge \pink \: solution$

$\bf \large taking \: y = {x}^{ sinx } .by \: taking \: logarithm \: on \: both \: sides \: we \: have \: \\ \\ \sf log \: (y) = \sin(x) log(y) \\ \\ \sf \: \frac{1}{y} . \frac{dy}{dx} = \sin(x \frac{d}{dx} ) ( log \: x) + log(x \frac{d}{dx} ) ( \sin \: x ) \\ \\ \sf or \ \: \: \: \frac{1}{y} \frac{dy}{dx} = ( \sin \: x ) \frac{1}{x} + log \: x \: cos \: x \\ \\ \sf or \: \: \frac{dy}{dx} = y ( \frac{sin \: x}{x} + \cos(x) log(x) \\ \\ \sf = {x}^{ \sin(x) } ( \frac{ \sin(x) }{x} + \cos(x) log(x) ) \\ \\ \sf = {. x}^{ sin \: x - 1} . \sin(x) + {x}^{ \sin(x) } .cos \: x \: log \: x$

Hope it's help u