Whats the derivative forumla for cos(u)
Doing -sin(u) doesnt seem to work.
For eg. Derivative of cos(-2x)
There is a subtlety about differentiation that isn't always stressed when teaching it.
Consider the equation
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?<br />\frac{d}{dx}\sin{(x)} = \cos{(x)}<br />)
Here, x appears three times: once in the denominator, and once as the argument of sine and cosine.
Why does this next equation fail?
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?<br />\frac{d}{dx}\sin{(2x)} = \cos{(2x)}<br />)
It's because it doesn't match the pattern above. We've swapped x for 2x in two out of three places. The denominator still says x, however. If you use a chain rule and sub in u=2x, then the derivative you're looking for then looks like the above pattern, with the three x's.
So, when calculating something like
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?<br />\frac{d}{dx}\sin{(\cos{(x)})}<br />)
the first step is to make this look something like the derivative rule for sine. However, the argument of the sine function here, which is a cosine, isn't x, so we have to use a chain rule. Let u = cos x. Then, by the chain rule,
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?<br />\frac{d}{dx}\sin{(\cos{(x)})} = \frac{du}{dx}\frac{d}{du}\sin{(u)}= -\sin{(x)} \cos{(u)}= -\sin{(x)}\cos{(\sin{(x)})}<br />)
It's important to understand what the chain rule does. It essentially changes the derivative from d/dx to d/du, which then allows you to apply the familiar derivative rules. THIS is why we have a chain rule.