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

Question

Samantha Adkison · Accepted Answer

#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.

Scarlett Allison · Answer

undefined To differentiate $ f(x) = \frac{x^2}{\ln(\tan(x))} $ using the quotient rule:

1. Let $ u(x) = x^2 $ and $ v(x) = \ln(\tan(x)) $.
2. Apply the quotient rule: $ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $.
3. Find the derivatives:
   - $ u'(x) = 2x $ (derivative of $ x^2 $).
   - $ v'(x) = \frac{1}{\tan(x)} \cdot \sec^2(x) = \frac{\sec^2(x)}{\tan(x)} $ (derivative of $ \ln(\tan(x)) $).
4. Substitute into the quotient rule formula:
   $$ f'(x) = \frac{(2x) \cdot \ln(\tan(x)) - x^2 \cdot \frac{\sec^2(x)}{\tan(x)}}{(\ln(\tan(x)))^2} $$
5. Simplify if needed.