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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?

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Have you learned the Chain Rule? –  Amzoti Jan 20 '13 at 19:48

2 Answers 2

up vote 1 down vote accepted

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

Solution:

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 $u = nx$:

$$ -n \sin nx $$

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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 : $$(\cos nx)'=\frac{d(\cos(u))}{du}\times\frac{du}{dx}=-\sin(u)\times x=-\sin(nx)\times x$$

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

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