Derivatives of trig functions

How can I prove that $\frac{d}{dx} (\csc x)= -\csc x \cot x$?

Specifically, how does one see the step $\cos x/\sin x = \cot x$?

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$\frac{\cos(x)}{\sin(x)}$ is the definition of cotangent. – EuYu Oct 23 '12 at 19:03
Depends how $\cot$ is defined. If it is defined as $\frac{1}{\tan x}$, then you just want the reciprocal of $\frac{\sin x}{\cos x}$. – André Nicolas Oct 23 '12 at 19:04

The derivative of $\csc(x)$ is the same thing as the derivative of $1/\sin(x)$. Naturally you can bring the $\sin(x)$ up to the front with a negative exponent like this: $\sin(x)^{-1}$. Using the chain rule you find the derivative of the outer ($u^{-1}$) and the inner ($\sin(x)$). You should know the power rule and that the derivative of $\sin(x)$ is $\cos(x)$. Multiply the two together to get $-\sin^{-2}(x) \cos(x)$. To make the next step easier to understand, put the $\sin^2(x)$ under $-\cos(x)$. Since $\sin^2(x)$ is the same as $(\sin(x))^2$ you can split it up to look like this: $$-\frac{1}{\sin(x)} \cdot \frac{\cos(x)}{\sin(x)}.$$ Naturally $1/\sin(x)$ becomes $\csc(x)$ and $\cos(x)/\sin(x)$ is $\cot(x)$. After simplifying you get that the derivative of $\csc(x)$ is actually $-\csc(x)\cot(x)$.

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There are a few ways to do this. I'm going to assume you know the chain rule and how to differentiate sine and cosine. Then,

$$\frac{d}{dx} \csc(x) = \frac{d}{dx} \frac{1}{\sin(x)} = \frac{d}{dx} (\sin(x))^{-1}$$

At this point, use the chain rule. You know the derivative of $u^{-1}$ is $- u^{-2}$, and also that the derivative of sine is cosine. Then, we have

$$\frac{d}{dx} \csc(x) = -(\sin(x))^{-2} \cos(x) = -\frac{1}{\sin(x)} \frac{\cos(x)}{\sin(x)} = -\csc(x) \cot(x)$$.

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