How can I prove that $\frac{d}{dx} (\csc x)= -\csc x \cot x$?
Specifically, how does one see the step $\cos x/\sin x = \cot x$?
[Math] Derivatives of trig functions
calculusderivativestrigonometry
Related Solutions
First, you should be writing $\frac{d}{dx}$, not $\frac{dy}{dx}$. $\frac{dy}{dx}$ refers to the derivative of $y$ with respect to $x$, while here you are taking the derivative of some complicated function with respect to $x$. After that, this is just an application of the chain rule. On the right-hand side, $$\frac{d}{dx}(-2y) = -2\frac{dy}{dx} = -2y'(x).$$ On the left-hand side, \begin{align} \frac{d}{dx}[\cos\cos(x^3y^2) - x \cot y] &= \frac{d}{dx}(\cos\cos(x^3y^2)) - \frac{d}{dx}(x\cot y) \\ &= -\sin\cos(x^3y^2)\cdot\frac{d}{dx}(\cos(x^3y^2)) - \cot y - x\frac{d}{dx}(\cot y) \\ &= -\sin\cos(x^3y^2)\left(-\sin(x^3y^2)\right)\cdot\frac{d}{dx}(x^3y^2) - \cot y + x\csc^2 y\cdot\frac{dy}{dx} \\ &= \sin\cos(x^3y^2)\sin(x^3y^2)\left(3x^2y^2 + 2x^3y\frac{dy}{dx}\right) - \cot y + x\csc^2 y\cdot\frac{dy}{dx} \\ &= \sin\cos(x^3y^2)\sin(x^3y^2)(3x^2y^2 + 2x^3yy') - \cot y + xy'\csc^2 y. \end{align} Set those two equal and solve for $y'$.
Yes, there are nice geometric explanations of the derivative formulas for all six basic trig functions, which ought to be much more widely known. (I hesitate to use the word "proof" for an argument that uses infinitesimals.) They are all based on the following fact about isosceles triangles:
For an isosceles triangle with small vertex angle $d\theta$, the base length $ds$ satisfies $$ ds \;\approx\; r\,d\theta $$ where $r$ is the length of the legs.
As you already pointed out, there is a nice geometric explanation of the derivative formulas for $\sin \theta$ and $\cos \theta$ that uses this fact. The following picture shows a right triangle version of the explanation, as opposed to the unit circle version you gave above:
The following picture shows a geometric explanation of the derivative formulas for $\sec \theta$ and $\tan \theta$, again using right triangles.
Finally, the following picture shows a geometric explanation of the derivative formulas for $\csc \theta$ and $\cot \theta$.
Best Answer
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
$$\frac{d}{dx} \csc(x) = -(\sin(x))^{-2} \cos(x) = -\frac{1}{\sin(x)} \frac{\cos(x)}{\sin(x)} = -\csc(x) \cot(x)$$.