What is the derivative of xsin⁡x

${\displaystyle \frac{dy}{dx}=x\cos x+\sin x}$
Explanation:
We have:
${\displaystyle y=x\sin x}$
Which is the product of two functions, and so we apply the Product Rule for Differentiation:
${\displaystyle \frac{d}{dx}\left(uv\right)=u\frac{dv}{dx}+\frac{du}{dx}v,\text{ or }{\left(uv\right)}^{\prime }=\left(du\right)v+u\left(dv\right)}$
I was taught to remember the rule in words; "The first times the derivative of the second plus the derivative of the first times the second ".
So with ${\displaystyle y=x\sin x}$
$$\{\begin{array}{lll}\text{Let}& u=x& \Rightarrow \frac{du}{dx}=1\\ \text{And }& v=\sin x& \Rightarrow \frac{dv}{dx}=\cos x\end{array}$$
Then:
${\displaystyle \frac{d}{dx}\left(uv\right)=u\frac{dv}{dx}+\frac{du}{dx}v}$
Gives us:
${\displaystyle \frac{d}{dx}\left(x\sin x\right)=\left(x\right)\left(\cos x\right)+\left(1\right)\left(\sin x\right)}$
${\displaystyle \therefore \frac{dy}{dx}=x\cos x+\sin x}$
If you are new to Calculus then explicitly substituting u and v can be quite helpful, but with practice these steps can be omitted, and the product rule can be applied as we write out the solution.

This is a function which is in the form,
${\displaystyle y=f\left(x\right)g\left(x\right)}$
Its

Finally found the answer to this difficult question, I'm happy