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

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Answer 1 · 2016-05-18T14:09:51

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

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.