Question

What is the derivative of `sec^-3(x)`?

Answer 1

`(dy)/(dx)=-3sinxcos^2x`

Explanation:

Here ,

`y=sec^(-3)(x)=1/sec^3x=cos^3x`

`"Using"color(blue)" Chain Rule :"`

`(dy)/(dx)=3cos^2xd/(dx)(cosx)`

`:.(dy)/(dx)=3cos^2x(-sinx)=-3sinxcos^2x`

Answer 2

`1/3\cos^3(\sec^{-3}x)\cot (sec^{-3}x)`

Explanation:

Given function:

`y=\sec^{-3}(x)`

`x=\sec^3y`

Now, differentiating above equation w.r.t. on both the sides by using chain rule as follows

`d/dx(x)=d/dx(\sec^3y)`

`1=3\sec^2yd/dx(\sec y)`

`1=3\sec^2y(sec y\tan y)dy/dx`

`1=3\sec^3y\tan ydy/dx`

`dy/dx=1/3\cos^3y\cot y`

`dy/dx=1/3\cos^3(\sec^{-3}x)\cot (sec^{-3}x)`

Answer 3

`-3cos^2(x)sin(x)`

Explanation:

We have `sec(x)=(cos(x))^(-1)` so `(sec(x))^(-3)=cos(x)^3`
so we get by the chain rule

`(cos(x)^3)'=3cos(x)^2(-sin(x))=-3cos^2(x)sin(x)`