How do you differentiate # f(x) =1- (sec x/ tan x)^2 #?

2 Answers
May 30, 2018

#\frac{2cos(x)}{sin^3(x)}#

Explanation:

By definition, #sec(x)= \frac{1}{cos(x)}#, and #tan(x) = \frac{sin(x)}{cos(x)}#

We can rewrite the fraction in parenthesis as

#\frac{sec(x)}{tan(x)} = \frac{1}{cos(x)}\cdot \frac{cos(x)}{sin(x)} = \frac{1}{sin(x)}#

So, the expression becomes

#1 - \frac{1}{sin^2(x)}#

To differentiate this expression, remember that

#d/(dx) (1 - \frac{1}{sin^2(x)}) = d/(dx) (1) - d/(dx) (\frac{1}{sin^2(x)}) =- d/(dx) \frac{1}{sin^2(x)}#

since the derivative of a number is zero.

Finally, you can write #-1/sin^2(x)# as #-sin^{-2}(x)#, and derive it with chain rule:

#d/(dx) -sin^{-2}(x) = -d/(dx) sin^{-2}(x) = - (-2sin^{-3}(x)*cos(x)) = \frac{2cos(x)}{sin^3(x)}#

#f'(1-csc^2x)=2csc^2xxxcotx#

Explanation:

#f(x)=1-(secx/tanx)^2#

#secx=1/cosx#
#tanx=sinx/cosx#

#secx/tanx=(1/cosx)/(sinx/cosx)#

Multiplying numerator and denominator by cosx

#secx/tanx=1/sinx#
#(secx/tanx)^2=(1/sinx)^2=1/sin^2x#

#1-(secx/tanx)^2=1-(1/sinx)^2#
#(1/sin^2x)=csc^2x#
#1-(secx/tanx)^2=1-csc^2x#

#f(x)=1-csc^2x#

#f'(x)=f'(1-csc^2x)#

#=f'(1)-f'(csc^2x)#

#f'(1)=0#
#f'(csc2x)=2cscxxx(-cscxxxcotx)#

#f'(csc^2x)=2csc^2xxxcotx#

#f'(1-csc^2x)=0-(-2csc^2xxxcotx)#

#f'(1-csc^2x)=2csc^2xxxcotx#