Question

How do you differentiate `y=3^(cotx)`?

Answer 1

This is an interesting problem.

There are different strategies of approaching this question, but I will take what I see simplest: use of the chain rule.

Let `y = 3^u` and `u = cotx`. We have to differentiate both.

`y = 3^u`

`ln(y) = ln(3^u)`

`lny = u(ln3)`

`1/y(dy/(du)) = ln3`

`dy/du = yln3`

`dy/(du) = 3^(u)ln3`

Now for `u`:

`u = cosx/sinx`

By the quotient rule:

`u' = (-sinx xx sinx - cosx xx cosx)/(sinx)^2`

`u' = (-sin^2x - cos^2x)/(sin^2x)`

`u' = (-(sin^2x + cos^2x))/(sin^2x)`

`u' = (-1)/sin^2x`

`u' = -csc^2x`

By the chain rule:

`dy/dx = 3^(u)ln3 xx -csc^2x`

`dy/dx = -csc^2x xx 3^(cotx)xxln3`

In summary, the derivative of `y = 3^(cot(x))` is `dy/dx = -csc^2x xx 3^(cotx)xxln3`.

Hopefully this helps!