Question

If y=e^x sinx show that d^2y/dx^2 = 2e^x cosx

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

$\displaystyle \sf \longrightarrow \frac{d^2y}{dx^2}=2.e^x.\cos x \ [ \ Shown \ ]$

Step-by-step explanation:

Given :

$\displaystyle \sf y=e^{x} \sin x$

We have to show :

$\displaystyle \sf \frac{d^2y}{dx^2} =2 \ e^x \cos x$

We know :

$\displaystyle \sf \frac{d}{dx} (e^x)=e^x$

$\displaystyle \sf \frac{d}{dx} (\sin x)=\cos x$

$\displaystyle \sf \frac{d}{dx} (\cos x)=\sin x$

Diff. given function w.r.t. x :

$\displaystyle \sf \frac{dy}{dx}= y'=\frac{d}{dx}\left(e^{x}. \sin x \right)$

Using product rule :

$\displaystyle \sf \frac{d}{dx}\left(f(x).g(x) \right)=g(x).f'(x)+f(x).g'(x)$

$\displaystyle \sf \longrightarrow \frac{dy}{dx}=\frac{d}{dx}\left(e^{x}. \sin x \right)$

$\displaystyle \sf \longrightarrow \frac{dy}{dx}= e^{x}. (\sin x)'+\sin x.(e^{x})'$

$\displaystyle \sf \longrightarrow \frac{dy}{dx}= e^{x}.\cos x+e^{x} .\sin x$

Now finding second order derivative :

$\displaystyle \sf \longrightarrow \frac{d^2y}{dx^2}= \frac{d}{dx} (e^{x}.\cos x+e^{x} .\sin x)$

$\displaystyle \sf \longrightarrow \frac{d^2y}{dx^2}= \frac{d}{dx} (e^{x}.\cos x)+\frac{d}{dx} (e^{x} .\sin x)$

Again using product rule :

$\displaystyle \sf \longrightarrow \frac{d^2y}{dx^2}=e^{x}.(\cos x)'+\cos x.(e^x)'+ e^{x} .(\sin x)'+\sin x.(e^x)'$

$\displaystyle \sf \longrightarrow \frac{d^2y}{dx^2}=e^{x}.(-\sin x)+\cos x.e^x+ e^{x} .(\cos x)+\sin x.e^x$

$\displaystyle \sf \longrightarrow \frac{d^2y}{dx^2}=e^x.\cos x+ e^{x} .\cos x$

$\displaystyle \sf \longrightarrow \frac{d^2y}{dx^2}=2.e^x.\cos x$

Hence shown.