Applying chain rule in $f(x)=\sin(x)\cdot x\ln(x)$ 
Can we apply  chain rule in function


$f(x)= \sin(x)\cdot x\ln(x)$

What i try:: Chain rule
$$\frac{d}{dx}\bigg(f(g(x)\bigg)=f'(g(x))\cdot g'(x)$$
So $$\frac{d}{dx}\bigg(\sin (x)\cdot x\ln(x)\bigg)=\sin(x)\cdot \frac{d}{dx}\bigg(x\ln(x)\bigg)+x\ln(x)\frac{d}{dx}(\sin x)$$
I have seems that we can not apply chain rule Here. Please conform me whether i am right or not. Thanks
 A: You are correct, this is just applying the product rule twice. What is the derivative of $x\ln(x)$?
A: The chain rule is applied when you have compositions of functions. There's no such occurrence here. What you have is products of functions. Hence, you have to apply the product rule.
Using the product rule we have:
$$\frac{d}{dx}\bigg(\sin (x)\cdot x\ln(x)\bigg)=\sin(x)\cdot \frac{d}{dx}\bigg(x\ln(x)\bigg)+\frac{d}{dx}(\sin x)\cdot x\ln(x)$$
$$=\sin(x)\cdot \frac{d}{dx}\bigg(x\ln(x)\bigg)+\cos(x)\cdot x\ln(x)$$
Using the product rule again we have:
$$\frac{d}{dx}\bigg(x\ln(x)\bigg)=x\cdot\frac{d}{dx}\bigg(ln(x)\bigg)+\frac{d}{dx}\bigg(x\bigg)\cdot ln(x)$$
$$\frac{d}{dx}\bigg(x\ln(x)\bigg)=x\cdot\frac{1}{x}+ln(x)=ln(x)+1$$
By substituting in the first equation we get:
$$\frac{d}{dx}\bigg(\sin (x)\cdot x\ln(x)\bigg)=\sin(x)\cdot(ln(x)+1)+cos(x)\cdot x\cdot ln(x)$$
A: Forgot the product rule ? I suggest that you use that rule there in your question. The reply by @ravindra-ranwala is all what I have to say. Even @phil said it right..
