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I've just started high school calculus. To differentiate trig functions the rule is $(f \circ g)' = g'(x) \cdot f'(g(x))$

So for $\tan(-x)$ would this not be $-\sec^2(-x)$? The answer says $-\sec^2(x)$ which is confusing as this was the only one I got incorrect.

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sec^2(-x) = sec^2(x). – Alex Zorn Apr 2 '14 at 20:36
Complain fiercely! ;-) – egreg Apr 2 '14 at 20:47
up vote 2 down vote accepted

Note that $$\sec(\theta)=\frac{1}{\cos(\theta)}$$ and $$\cos(-\theta)=\cos(\theta)$$ So your answer is correct, really.

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Recall that $$\sec(x) = \frac{1}{\cos(x)}.$$

Cosine is an even function: $\cos(-x) = \cos(x)$. What does that say about $\sec(-x)$? What about $-\sec^2(-x)$?

Incidentally, $-\sec^2(-x)$ is not any "less correct" than $-\sec^2(x)$ or $\frac{-1}{\cos^2(x)}$ or any number of other possible ways of writing the answer. Unless your teacher gave you specific instructions for how to write your answers in some canonical simplest form, it might be worth asking about getting some points back if yours was marked wrong.

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