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If y = ex cos x, prove that \(\frac{d^2y}{dx^2}=2e^xcos(x+\frac{\pi}{2})\)

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 Basic Idea: Second order derivative is nothing but derivative of derivative i.e. \(\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})\)

The idea of chain rule of differentiation: If f is any real-valued function which is the composition of two functions u and v, i.e. f = v(u(x)). For the sake of simplicity just assume t = u(x)

Then f = v(t). By chain rule, we can write the derivative of f w.r.t to x as:

Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..

Let’s solve now:

Given,

y=excos x

TO prove :

\(\frac{d^2y}{dx^2}=2e^xcos(x+\frac{\pi}{2})\) 

Clearly from the expression to be proved we can easily observe that we need to just find the second derivative of given function.

Given, y = ex cos x

 As, \(\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})\)

So lets first find dy/dx and differentiate it again.

\(\therefore \frac{dy}{dx}=\frac{d}{dx}(e^xcosx)\)

Let u = ex and v = cos x

As, y = u*v

∴ Using product rule of differentiation:

Again differentiating w.r.t x:

Again using the product rule :

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