Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
sec^2(-x) = sec^2(x). –  Alex Zorn Apr 2 at 20:36
    
Complain fiercely! ;-) –  egreg Apr 2 at 20:47

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.