Question

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

Answer 1

`\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)}`

Answer 2

`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`