Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to calculate derivative of $ \cos ax$? Do I need any formula for $ \cos ax$?

The answer in my exercise book says it is $-a \sin ax$. But I don't know how to come to this result. Could you maybe explain it to me?

share|cite|improve this question
Have you learned the Chain Rule? – Amzoti Jan 20 '13 at 19:48
up vote 1 down vote accepted

Hint: Apply the Chain Rule with $u=nx$.


We have: $$ u = nx $$ Applying the chain rule: $$ \frac{\mathrm dy}{\mathrm dx} = \frac{\mathrm dy}{\mathrm du} \cdot \frac{\mathrm du}{\mathrm dx} $$

It is clear that $u' = n$. We now have: $$ \left(\frac{\mathrm d}{\mathrm du} \cos u\right) \cdot n $$ $$ -n \sin u $$

Substituting back $u = nx$:

$$ -n \sin nx $$

share|cite|improve this answer

As @Amzoti and @George noted, you can use the Chain rule. If $f(u)=y$ and $u=g(x)$ then: $$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$$, so in your problem we have : $$(\cos nx)'=\frac{d(\cos(u))}{du}\times\frac{du}{dx}=-\sin(u)\times n$$ This is what you are supposed to get during another answer. What I want to add you here is that, you should always care about the variable and about the constants. For example if you want the derivative of $\cos nx$ such that $n$ is variable and $x$ is constant then we will have : $$(\cos nx)'=\frac{d(\cos(u))}{du}\times\frac{du}{dx}=-\sin(u)\times x=-\sin(nx)\times x$$

share|cite|improve this answer
Nice point, Babak! – amWhy Jan 20 '13 at 20:05
@amWhy: Thanks for your kind words. ;-) – Babak S. Jan 20 '13 at 20:07

Your Answer


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.