# Chain Rule applied to Trig Functions

Given $f(x)= \sin(\pi x)^{2}$, find the derivative.

Using the chain rule my work is as follows: $(\sin(\pi x)^2)'$ becomes $$2 \sin(\pi x) \cdot \frac{d}{dx}(\sin(\pi x)$$ The derivative of sin is cos, thus $$2 \sin(\pi x) \cdot \cos(\pi x) \cdot \frac{d}{dx}(\pi x)$$ The derivative of $\pi x$ is $\pi$, and the equation stretches to

$$2 \sin(\pi x) \cos(\pi x) \pi == 2 \pi \sin(\pi x) \cos(\pi x)$$

However, the book states the answer as $$2 \pi^{2}~ x ~\cos(\pi x)^{2}$$ and that definitely doesn't match my result. Where did I go wrong?

EDIT

Thanks to Arturo Madigan, Jonas Meyer, et al for their help.

I re-did the problem based on having $(\pi x)^{2}$, having the exponent rather than the sin function, and it seems I have a missing exponent as well.

Differentiating the terms of the function via the chain rule, I get $$(\pi x)^{2} [\frac{d}{dx}sin] \cdot \frac{d}{dx}(\pi x)^{2}$$

$$cos(\pi x)^{2} \cdot 2\pi x= 2~\pi~ x cos~(\pi x)^{2}$$

According to the book answer, $2~\pi x$ should actually be $2~ \pi^{2} x$

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I don't think you have given the correct function. Was the original function $f(x)=\sin((\pi x)^2)$? It may have been written as $f(x)=\sin(\pi x)^2$, which is ambiguous, but you seem to have misinterpreted it as $(\sin(\pi x))^2=\sin^2(\pi x)$, and you wrote $\sin(\pi x)$. Will you please clarify? –  Jonas Meyer Mar 7 '11 at 5:27
How did you go from $f(x)= sin(\pi x)$ to (it appears) $f'(x)=2 sin(\pi x) \cdot \frac{d}{dx}(sin(\pi x))$? The $2\sin(\pi x)$ appeared out of nowhere and there are mismatched parentheses. But Jonas Meyer may have a better guess. –  Ross Millikan Mar 7 '11 at 5:29
@Ross Millikan, $2 sin(\pi x)$ is the derivative of $sin(\pi x)$ in the first stop using the chain rule. My work posted shows my attempt at going into the function. @Jonas Meyer, I wrote the function as it appears in the book. There is only one set of parenthesis, and the exponent is outside it. –  Jason Mar 7 '11 at 5:34
@Jonas Meyer, its been edited. My mistake. –  Jason Mar 7 '11 at 5:38
@Jason: Thank you. The missing exponent was the reason for PEV's and Dactyl's answers. –  Jonas Meyer Mar 7 '11 at 5:41

Given the answer, the question was for the derivative of $\sin\Bigl( (\pi x)^2\Bigr)$; instead, you computed the derivative of $\Bigl(\sin(\pi x)\Bigr)^2$.

If you were computing the derivative of the latter, then your computations are correct; the derivative is $2\pi\sin(\pi x)\cos(\pi x)$.

But if you were asked for the derivative of $\sin\Bigl((\pi x)^2\Bigr)$, then of course you were looking at the wrong function, and that's why the answers don't match.

It's possible you had "$\sin(\pi x)^2$" and interpreted this as $(\sin(\pi x))^2$; usually, $\sin^2(\pi x)$ is used for the latter, so "$\sin(\pi x)^2$" would be interpreted as $\sin\Bigl((\pi x)^2\Bigr)$.

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+1 Great detective work. Should not be required, but I am not sure whether to complain to OP, his textbook, or the math establishment. –  Ross Millikan Mar 7 '11 at 5:40
but how do you know which to compute for? The book does give examples of how parenthesis placement can affect computation but $sin x^{2}$ can also be written as $sin^{2} x$, which adds to my confusion –  Jason Mar 7 '11 at 5:40
@Ross Millikan, I'd definitely say the math prof & book play a large part, but I also have to factor in my lack of sleep studying for the exam. The prof is a great guy, but its very hard to follow along with what he's saying in the class, which is why I have my laptop running the whole time so I can double-check what he's talking about. –  Jason Mar 7 '11 at 5:43
@Jason: Since I would use $\sin^2(\pi x)$ for the function $\Bigl(\sin (\pi x)\Bigr)^2$, then it seems reasonable that the other notation, $\sin (\pi x)^2$, is being used to mean something else, namely $\sin\Bigl( (\pi x)^2\Bigr)$. It's bad use of notation on the part of the book (there should be that second set of parentheses to make everything clear), but it's a process of elimination ("if they had meant x they would have written it this way; they didn't write it this way, so they must have meant y instead"). –  Arturo Magidin Mar 7 '11 at 5:44
I'm reminded of a sign that was supposedly posted in a newsroom (back in the day of lead type)-"the composing room has an infinite supply of periods to terminate short, complete sentences". Maybe that applies to parentheses, too. We got lots of 'em, so use 'em liberally. –  Ross Millikan Mar 7 '11 at 5:51

The derivative is $\pi \cos(\pi x)$.

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You're right, but why does the sin cancel out in the book answer? –  Jason Mar 7 '11 at 5:21
Based on the OP's steps and the answer in the back of the book, I'm pretty sure $\sin(\pi x)$ was not intended. (I elaborate in my comment on the question.) –  Jonas Meyer Mar 7 '11 at 5:28

Think of the function $\sin(\pi x)$ like this: $x \to \pi x \to \sin(\pi x)$. In which case it is intuitively clear that the rate of change of the function should be the multiplication of the rate of changes of $x \to \pi x$ and $y \to \sin (y)$, the latter being evaluated at $y=x$. Hence PEV's answer.

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