Differentiate the following function with respect of `x :(x^3+x^2+1)sinx`

Answer

`frac{d y}{d x}=(x^{3}+x^{2}+1) cos x+(3 x^{2}+2 x) sin x`

Explanation

Since y is a product of u and v where:

`u=x^3+x^{2}+1`
and `v=sin x`

therefore `y=uv`.

Now, we apply product rule of differentiation.

By product rule, we have -

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

As,`u=x^{3}+x^{2}+1`

...