Get the solution to "If y=xsinx, then dydx=?A) xsinxxcosxlogx+sinxxBxsinxxcosxlogx+cosxxC) yxcosxlogx+cosxxD) None of..."

Alaina Durham

Answered question

2022-12-21

y = x^{sinx},

then

\frac{dy}{dx} = ?

\frac{x^{\sin x} (x \cos x \log x + \sin x)}{x}

\frac{x^{sinx} (xcosxlogx + \cos x)}{x}

\frac{y (xcosxlogx + cosx)}{x}

D) None of these

mizzxrizzeui

Beginner2022-12-22Added 5 answers

The correct answer is A

\frac{x^{\sin x} (x \cos x \log x + \sin x)}{x}

Finding the value of

\frac{dy}{dx}

:
The given differential equation is

y = x^{sinx}

y = x^{sinx} \log y = \log x^{sinx} \log y = sinx \log x \frac{d}{dx} (\log y) = \frac{d}{dx} (sinx \log x) \frac{1}{y} \frac{dy}{dx} = sinx \frac{d}{dx} (\log x) + (\log x) \frac{d}{dx} (\sin x) \frac{dy}{dx} = y (\frac{1}{x} sinx + cosx logx) \frac{dy}{dx} = x^{sinx} (\frac{sinx}{x} + (cosx) \log x) \frac{dy}{dx} = \frac{x^{sinx} (xcosx \log x + sinx)}{x}

Therefore, the correct answer is (A).

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