Question

How do you differentiate `f(x)=(sinx+1)(x^2-3e^x)` using the product rule?

Answer 1

`\cos (x)(x^2-3e^x)+(2x-3e^x)(\sin (x)+1)`

Explanation:

`\frac{d}{dx}((\sin (x)+1)(x^2-3e^x))`

Applying product rule,
`(f\cdot g)^'=f^'\cdot g+f\cdot g^'`

`f=sinx +1 ,g=x^2 - 3e^x`
`=frac{d}{dx}(sin (x)+1)(x^2-3e^x)+frac{d}{dx}(x^2-3e^x)(sin (x)+1)`

we know,
`\frac{d}{dx}(sin (x)+1)=cos (x)`;
`\frac{d}{dx}(\sin (x))=\cos(x)`;
`\frac{d}{dx}(1)=0`;
`\frac{d}{dx}(x^2-3e^x)=2x-3e^x`

Finally,
`=\cos(x)(x^2-3e^x)+(2x-3e^x)(\sin (x)+1)`