Differentiate the following function with respect of `x :(2x^2-3)sin x`

Answer

`frac{d y}{d x}=(2 x^{2}-3) cos x+4 x sin x`

Explanation

Since y is the product of u and v where:

`u=2 x^{2}-3` and `v=\sin x`

So,` y=u v`

We will apply the product rule of differentiation.

`\frac{d y}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}` -equation 1

As, `u=2 x^{2}-3`

` \frac{du}{dx}=4 x` -equation 2

As, `v=\sin x`

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