Math, asked by aditya123215, 7 months ago

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

Answers

Answered by VinayGulati
0

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

Answered by CottenCandy
7

 \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

Similar questions