Question

What is the derivative of `[ln(2-3x)/lnx]^[arccos(2-5x)]`?

Answer 1

`y^' = [ln(2-3)/lnx]^(arccos(2-5x)) * [5/sqrt(1-(2-5x)^2) * ln(ln(2-3x)/lnx) + 1/ln(2-3x) * (-3/(2-3x) - ln(2-3x)/(x * lnx)) * arccos(2-5x)]`

Explanation:

!! EXTREMELY LONG ANSWER !!

The most important derivation rule you'll need to use to differentiate this function is the power rule for a variable base and a variable power.

`color(blue)(d/dx(f(x)^g(x)) = f(x)^g(x) * d/dx[ln(f(x)) * g(x)])`

In your case, you have `f(x) = ln(2-3x)/ln(x)` and `g(x) = arccos(2-5x)`. Now, I will assume that you know the derivative of `arccosx`

`d/dx(arccosx) = -1/(sqrt(1-x^2)`

Ok, buckle up because the calculations will be `color(red)("horrendous")`.

So, start your calculation by writing the derivative of `y = f(x)^g(x)` using the power rule. To keep the calculations as compact as possible, I'll use `f(x) =f` and `g(x) = g` from this point on.

`d/dx(y) = f^g * d/dx(ln(f) * g)" "color(purple)((1))`

Use the product rule to find

`d/dx(ln(f) * g) = [d/dxln(f)] * g + ln(f) * d/dx(g)" "color(purple)((2))`

Now break this calculation into two different ones. The first one will be

`d/dxln(f) = d/dx[ln(ln(2-3x)/lnx)]`

Use the quotient rule once and the chain rule twice, once for `ln(u)`, with `u = ln(2-3x)/lnx`, and once more for `ln(v)`, with `v = 2-3x)`.

`d/dx(ln(u)) = d/(du)ln(u) * d/dx(u) " "color(purple)((3))`

`d/dx(u) = ([d/dxln(2-3x)] * lnx - ln(2-3x) * d/dx(lnx))/(lnx)^2 " "color(purple)((4))`

Use the second chain rule substitution to get

`d/dx(lnv) = d/(dv)lnv * d/dx(v)`

`d/dx(lnv) = 1/v * d/dx(2-3x)`

`d/dx(ln(2-3x)) = 1/(2-3x) * (-3)`

Plug this back into `color(purple)((4))` to get

`d/dx(u) = (-3/(2-3x) * lnx - ln(2-3x) * 1/x)/ln^2x`

Now take this result back to `color(purple)((3))` to get

`d/dx(ln(u)) = 1/u * (-3/(2-3x) * lnx - ln(2-3x) * 1/x)/ln^2x`

`d/dx(ln(f)) = color(red)(cancel(color(black)(lnx)))/ln(2-3x) * (-3/(2-3x) * lnx - ln(2-3x) * 1/x)/ln^color(red)(cancel(color(black)(2)))x`

`d/dx(lnf) = 1/ln(2-3x) * (-3/(2-3x) - ln(2-3x)/(x * lnx))`

From `color(purple)((2))`, focus on finding `d/dx(g)`

`d/dx(g) = d/dx(arccos(2-5x))`

You're going to have to use th chain rule once for `arccost`, with `t = 2-5x)`

`d/dx(arccost) = d/(dt)arccost * d/dx(t)`

`d/dxarccost = -1/(sqrt(1-t^2)) * d/dx(2-5x)`

`d/dx(arccos(2-5x)) = -1/sqrt(1-(2-5x)^2) * (-5)`

`d/dx(g) = 5/sqrt(1-(2-5x)^2)`

Take everyting back to `color(purple)((2))` to get

`d/dx(ln(f) * g) = 1/ln(2-3x) * (-3/(2-3x) - ln(2-3x)/(x * lnx)) * arccos(2-5x) + ln(ln(2-3x)/lnx) * 5/sqrt(1-(2-5x)^2)`

Finally, take this back to `color(purple)((1))` to get

`y^' = color(green)([ln(2-3)/lnx]^(arccos(2-5x)) * [5/sqrt(1-(2-5x)^2) * ln(ln(2-3x)/lnx) + 1/ln(2-3x) * (-3/(2-3x) - ln(2-3x)/(x * lnx)) * arccos(2-5x)])`