Question

What is the derivative of `cot^2(sinx)`?

Answer 1

`(df)/dx = 2cot (sin x)((cos x)(-csc^2 (sin x))) = -2csc^2 (sin x)(cot (sin x))(cos (x))`

Explanation:

We will be using the chain rule in this solution. For finding the derivative of `cot^2 (u)`, we can either use the link to the proof at emathzone.com

http://www.emathzone.com/tutorials/calculus/derivative-of-cotangent-squared-function.html

OR we can find the derivative using the product rule.

Chain rule: given a function `f (x)=g (h (x)), (df)/(dx)=h'(x)*g'(h)`.

Product Rule: when asked to differentiate a function of the form `f (x)=g (x)*h (x), (df)/dx = g'(x)*h (x) + g (x)*h'(x)`

To use the product rule, we will state the problem in a different form:

`f (x)=cot (sin x)*cot (sin x)`.

Therefore...

`(df)/dx= (d/dx (cot (sin x)))cot (sin x) + cot (sin x)(d/dx (cot (sin x))) = 2cot (sin x)d/dxcot (sin x)`

We must use the chain rule to differentiate the second term here. Recall that `d/dx cot(x) = -csc^2x` (proof given at the following link for sake of brevity): http://www.math.com/tables/derivatives/more/trig.htm#reciprocals

Further, we know `d/dx sin x = cos x`. Therefore if our function is `cot (sin x)`, our derivative will take the form `cos (x)(-csc^2(sin x))`

Thus, our overal, derivative is;

`(df)/dx = 2cot (sin x)((cos x)(-csc^2 (sin x))) = -2csc^2 (sin x)(cot (sin x))(cos (x))`