Question

What is the derivative of `y=6(sec(x)-tan(x))^(3/2)`?

Answer 1

`dy/dx = 9(sec(x)-tan(x))^(1/2) (tan(x)sec(x)-sec^2(x))`

Explanation:

Given: `y=6(sec(x)-tan(x))^(3/2)`

Use the chain rule to compute the derivative:

`(d(y(u(x))))/dx = (d(y(u)))/(du) (du)/dx`

Let `u = sec(x)-tan(x)`, then:

`y = 6u^(3/2)`

`(d(y(u)))/(du) = 9u^(1/2)`

and

`(du)/dx = tan(x)sec(x)-sec^2(x)`

Substituting into the chain rule:

`dy/dx = 9u^(1/2) (tan(x)sec(x)-sec^2(x))`

Reversing the substitution for u:

`dy/dx = 9(sec(x)-tan(x))^(1/2) (tan(x)sec(x)-sec^2(x))`