Question

How do you differentiate `f(x)=x^2/ln(tanx)` using the quotient rule?

Answer 1

`f' (x)=(2x*ln(tan x)-x^2*csc x*sec x)/(ln(tan x))^2`

Explanation:

We start with the given function `f(x)=x^2/(ln(tan x))`

We use the quotient formula for finding derivatives

`d/dx(u/v)=(v*d/dx(u)-u*d/dx(v))/v^2`

Let `u=x^2` and `v=ln(tan x)`

Let us use the formula

`f' (x)=d/dx(x^2/ln(tan x))=(ln(tan x)*d/dx(x^2)-x^2*d/dx(ln(tan x)))/(ln(tan x))^2`

`f' (x)=(ln(tan x)*2x-x^2*1/(tan x)*d/dx(tan x))/(ln(tan x))^2`

`f' (x)=(ln(tan x)*2x-x^2*1/(tan x)*sec^2 x)/(ln(tan x))^2`

Take note that `1/tan x=cos x/sin x=csc x*cos x`

`f' (x)=(ln(tan x)*2x-x^2*csc x*cos x*sec^2 x)/(ln(tan x))^2`

`f' (x)=(ln(tan x)*2x-x^2*csc x*sec x)/(ln(tan x))^2`

final answer should be

`f' (x)=(2x*ln(tan x)-x^2*csc x*sec x)/(ln(tan x))^2`

God bless....I hope the explanation is useful.