How to implicitly differentiate $x \sin x = y(1 + \cos y)$?

Question

If $x \sin x = y(1 + \cos y)$, find $y'$ at $(\frac{\pi}{2}, \frac{\pi}{3})$. Please show each step.

Answer 1

Recall the following:

• Product rule: $\dfrac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$
• Chain rule: $\dfrac{d}{dx}[f(g(x))] = f'(g(x))g'(x)$
• $\dfrac{d}{dx}[\cos(x)] = -\sin(x)$
• $\dfrac{d}{dx}[\sin(x)] = \cos(x)$

When performing implicit differentiation, you decide which variable to differentiate. For instance, if we have an equation with variables $x$ and $y$, and we want to differentiate it with respect to $x$, then the derivative of $y$ is $y'$, whereas the $x$ terms are differentiated by the usual derivative rules.

For your problem, we have

$$\begin{aligned} x\sin(x) &= y(1 + \cos(y))\\ x\sin(x) &= y + y\cos(y) \end{aligned}$$

Then,

$$\begin{aligned} x\cos(x) + \sin(x) = y' + y'\cos(y) - y\sin(y) \cdot y' \end{aligned}$$

which is simplified as

$$y' = \dfrac{x\cos(x) + \sin(x)}{1 + \cos(y) - y\sin(y)}$$

Here is how this is derived:

1. The derivative of $x\cos(x)$ is found by product rule.
2. The derivative of $y$ with respect to $x$ is $y'$.
3. The derivative of $\cos(y)$ with respect to $x$ is $\sin(y)\cdot y'$. Use chain rule for this part.
4. To differentiate $y\cos(y)$ with respect to $x$, use product rule.

The rest is left for you to determine $y'$ for $\left(\frac{\pi}{2}, \frac{\pi}{3} \right)$. Simply substitute $x$ with $\frac{\pi}{2}$ and $y$ with $\frac{\pi}{3}$. Finally, solve for $y'$.

Answer 2

First, recall that $(\sin x)' = \cos x$, $(\cos x)' = -\sin x$, and $(\alpha)' = 0$ for any real number $\alpha$. Also, by the chain rule, the derivative of $f(y)$ with respect to $x$ is $f'(y)y'$.

If we differentiate both sides of $x \sin x = y(1 + \cos y)$ with respect to $x$, we get $$x \cos x + \sin x = -yy' \sin y + (1 + \cos y)y'.$$ Note that we used both the product rule and the chain rule. Hence $$x \cos x + \sin x = [(1 + \cos y) - y \sin y]y',$$ that is, $$y' = \frac{x \cos x + \sin x}{1 + \cos y - y \sin y}.$$ Now substitute $x = \pi/2$ and $y = \pi/3$.