Is applying chain rule in Trigonometric function wrong?

Question

Find $dy/dx$ if $y = x \sin x$.

I was trying to solve the problem following way.

$$\frac{d}{dx}[\sin x]=x\cos x$$

So,

$$\frac{d}{dx}[x \sin x]$$ $$=>x\frac{d}{dx}[\sin x]+\sin x \frac{d}{dx}[x]$$ $$=>x^2 \cos x+\sin x$$

But, when I saw answer in Calculus by Howard Anton book I saw my answer was wrong.

$$\frac{d}{dx}[x \sin x]$$ $$=>x\frac{d}{dx}[\sin x]+\sin x \frac{d}{dx}[x]$$ $$=>x \cos x+\sin x$$

Why the book answer didn't apply chain rule? So, is applying chain rule in Trigonometric function wrong?

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In my Physics, I found an equation

$$x=A \sin (\omega t)$$

When they differentiate x than, they wrote something just like this.

$$x=A\omega \cos (\omega t)$$

I just used chain in my main question from this method. $$\frac{d}{dx}[\sin x]=x \cos x$$

Actually, What am I thinking wrong here?

Answer 1

$ \frac{d}{dx}[\sin x] = \cos x$, not $ x\cos x$.

Your understanding of the Physics text is probably mistaken. In it, $\omega$ is a constant. The application of chain rule would go:

$$\frac{d}{dt}(A\sin(\omega t)) = A \cos (\omega t) \frac{d}{dt}(\omega t) = A\omega \cos (\omega t).$$

In contrast, the expression inside your sin term has no constant attached (or more precisely, that constant is 1, so it's basically invisible). Your expression is a special case of the above expression with $A=\omega=1$.

Answer 2

At the given case, the appropriate property to be applied is the product rule.

It says that if $f:\mathbb{R}\to\mathbb{R}$ and $g:\mathbb{R}\to\mathbb{R}$ are functions defined and differentiable at $x\in \mathbb{R}$, where $x$ is a limit point, then the following relation holds: \begin{align*} [f(x)g(x)]' = f'(x)g(x) + f(x)g'(x) \end{align*}

At your case, $f(x) = x$ and $g(x) = \sin(x)$.

Hopefully this helps!