Question

How do you differentiate `y=cot(1-2x^2)`?

Answer 1

`y^' = 4x * csc^2(1-2x^2)`

Explanation:

You can differentiate this function by using the chain rule for `cotu`, with `u = 1-2x^2`.

I'll assume that you don't know what the derivative of `cotx` is, but that you do know that

`color(blue)(d/dx(cosx) = -sinx)" "` and `" "color(blue)(d/dx(sinx) = cosx)`

You can derive the derivative of `cotx` by using the quotient rule or by using the product rule and the chain rule. You can find the derivative of `cotx` by using the quotient rule here, so I'll use the product + chain rules.

Start from the fact that

`cot(x) = cosx/sinx`

this can be rewritten as a product of two functions

`cotx = cosx * sin^(-1)x`

If you take `t^(-1)` with `t = sinx`, you can write

`d/dx(cotx) = [d/dx(cosx)] * sin^(-1)x + cosx * [d/dx(t^(-1))]`

`d/dx(cotx) = -color(red)(cancel(color(black)(sinx))) * color(red)(cancel(color(black)(sin^(-1)x))) + cosx * [d/(dt)t^(-1) * d/dx(t)]`

`d/dx(cotx) = -1 + cosx * [-t^(-2) * d/dx(sinx)]`

`d/dx(cotx) = -1 + cosx * (-1/sin^2x * cosx)`

`d/dx(cotx) = -1 -cos^2x/sin^2x`

This can be written as

`d/dx(cotx) = -(overbrace(sin^2x + cos^2x)^(color(blue)(=1)))/sin^2x = -1/sin^2x = -csc^2x`

Your original derivative will thus be

`d/dx(y) = d/(du)(cotu) * d/dx(u)`

`y^' = -csc^2u * d/dx(1-2x^2)`

`y^' = -csc^2(1-2x^2) * (-4x)`

Finally, you get that

`y^' = color(green)(4x * csc^2(1-2x^2))`