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