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

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Answer 1 · 2016-05-23T07:49:40

$cos (x) (x^{2} - 3 e^{x}) + (2 x - 3 e^{x}) (sin (x) + 1)$ $\cos (x)(x^2-3e^x)+(2x-3e^x)(\sin (x)+1)$

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

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

How do you differentiate f(x)=(sinx+1)(x^2-3e^x)f(x)=(sinx+1)(x2−3ex)f(x)=(sinx+1)(x^2-3e^x) using the product rule?