How do you differentiate f(x)=xsinx using the product rule?

hi
PRODUCT RULE SAYS
if to diffrentiate "uv"( here i will take w.r.t to x)
i.e
$d \frac{\left(u v\right)}{\mathrm{dx}}$=$u \cdot d \frac{v}{\mathrm{dx}}$+$v \cdot \frac{\mathrm{du}}{\mathrm{dx}}$
i.e u must know
diffrentiation of ${x}^{n}$=$n {x}^{n - 1}$
here..,

**diffrentiation of x=(1)${x}^{1 - 1}$ =1

diffrentiation of" sinx" is "$\cos x$"

NOW,
$\mathrm{dx} . \sin \frac{x}{\mathrm{dx}}$=$\left(x\right) . \mathrm{ds} \in \frac{x}{\mathrm{dx}}$+$\left(\sin x\right) \left(d \frac{x}{\mathrm{dx}}\right)$ = xcos(x)+sin(x)
=
So diffrentiation of F(X)=x sin(x)-=x cos(x)+sin(x)

Nov 12, 2017

$f ' \left(x\right) = x \cos x + \sin x$

Explanation:

$\text{given "f(x)=g(x)h(x)" then}$

$f ' \left(x\right) = g \left(x\right) h ' \left(x\right) + h \left(x\right) g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{product rule}}$

$g \left(x\right) = x \Rightarrow g ' \left(x\right) = 1$

$h \left(x\right) = \sin x \Rightarrow h ' \left(x\right) = \cos x$

$\Rightarrow f ' \left(x\right) = x \cos x + \sin x$