How do I use the chain rule with division?

Question

I've looked through YouTube and while I do believe I've found my answer to a question I'm confused over in regards to how to use the quotient rule with the chain rule, I'm still confused in terms of why my answer turned out different.

So, the question is:

$\left(\frac{\sin(x)}{1+\cos(x)}\right)^2$

My work is as follows:

$2\frac{\sin(x)}{1+\cos(x)} \times \frac{\sin(x)}{1+\cos(x)}$

Now, I know for "$\frac{\sin(x)}{1+\cos(x)} $" I got to find the derivative which involves me using the quotient rule.

So, I end up with (simplified): $\frac{2}{1+\cos(x)^2}$

Now, my final answer in terms of the chain rule is:

$2\frac{\sin(x)}{1+\cos(x)} \times \frac{2}{1+\cos(x)^2}$

Now after further simplifying, I end up with:

$\frac{4\sin(x)}{1+\cos^3(x)}$

I checked my answer in the back of my math textbook but they got the answer of:

$\frac{2\sin(x)}{(1+\cos(x))^2}$

I can see where they got it from, and I too would get the same answer if I did not include the quotient rule. But the thing is, it seems like they did use the quotient rule because otherwise, how could the denominator be squared?

Is my answer correct or incorrect? If so, what went wrong?

I'm following the chain rule formula to solve it.

I've also checked the graphs for my answer and the book's and the graphs are practically identical.

Thank you for your help!

Answer 1

Here is an outline of the general procedure. You can compare with yours to see where you may have gone wrong:

$$y=\left(\frac fg\right)^2$$ $$y' = 2\left(\frac fg\right)\cdot\left(\frac fg\right)'$$ $$= 2\left(\frac fg\right)\cdot\left( \frac{gf'-fg'}{g^2}\right)$$

In your case, $f(x) = \sin(x)$ and $g(x) = 1+\cos x$, so $f'(x) = \cos x$ and $g'(x) = -\sin x$. That means $$y' = 2\left( \frac{\sin x}{1 +\cos x}\right)\left(\frac{(1+\cos x)\cos x - \sin x(-\sin x)}{(1+ \cos x)^2} \right)$$ $$=\frac{2\sin x(\cos x + \cos^2 x + \sin^2 x)}{(1+\cos x)^3}$$ $$=\frac{2\sin x(\cos x + 1)}{(1+\cos x)^3}$$ $$=\frac{2\sin x}{(1+\cos x)^2}$$

Answer 2

The problem is that your result for the derivative of the fraction is wrong (and the book is indeed right).

Here is how I do it: \begin{align} \biggl(\frac{\sin x}{1+\cos x}\biggr)^{\kern-5mu'}&=\frac{\cos x(1+\cos x)-\sin x(-\sin x)}{(1+\cos x)^2}\\ &=\frac{\cos x+\overbrace{\cos^2 x+\sin^2 x}^{=1})}{(1+\cos x)^2}=\frac 1{1+\cos x}. \end{align}

Answer 3

For what it's worth, writing the original function as $\tan^2(x/2)$ would make the use of $(f^2)^\prime=2ff^\prime$ easier. So the derivative is$$\tan\tfrac{x}{2}\sec^2\tfrac{x}{2}=\frac{4\sin\tfrac{x}{2}\cos\tfrac{x}{2}}{(2\cos^2\tfrac{x}{2})^2}=\frac{2\sin x}{(1+\cos x) ^2}.$$