Question

How do you differentiate `y=csc^-1x-4cot^-1x`?

Answer 1

`dy/dx = -1/(|x|sqrt(x^2-1)) + 4/(1+x^2)`

Explanation:

Actually it isn't too bad!

Step I

Let `theta = csc^-1x`
`:. x = csc theta`

I shall assume that there is no need to show the step by step differentiation of `csc`, if I may (it just requires the chain rule, with `x= (sin theta)^-1`, and skip straight to:

`dx/(d theta) = -csc theta cot theta`

We see that `csc theta` is of course just `-x`. Then use the Pythagorean Identity to get `cot theta` in terms of `x`.

`1 + cot^2x = csc^2x`
`:.cot theta = sqrt (x^2-1)` (assume the positive square root here).

`:. dx/(d theta) = -xsqrt(x^2-1)`
`:. (d theta)/dx = -1/(xsqrt(x^2-1)`

Step II

Let `alpha = cot^-1x`
`:. x = cot alpha`

Again, if I may, I shall skip right to:

`dx / (d alpha) = -csc^2alpha`

Again, using the same Pythagorean Identity, we have:

`1 + x^2 = csc^2 alpha`
`:. -(1 + x^2) = dx/(d alpha)`
`:. (d alpha)/dx = -1/(1 +x^2)`

Step III

Now combine the lot:

`dy/dx = -1/(|x|sqrt(x^2-1)) + 4/(1+x^2)`, `x in (-oo,-1)uu(1, oo)`

Note: remember to re-introduce the constant `4` which I left out whilst differentiating `4cot^-1x`

Second note: the absolute value bars rather mysteriously appeared on the denominator of that derivative. I believe this is because actually sketching the graph of `y=csc^-1x`, the function is strictly decreasing, so by placing absolute value bars around the stray `x`, ensures that the derivative function stays negative for all `x`.

Third note: the domain comes in because the `csc` function is defined across this domain. The `cot` function is defined for all `x`, so the combined function takes the domain of the former.