Question

What is the derivative of `f(x)=1/(sinx+cosx)^2`?

Answer 1

`d/dx1/(sin(x) + cos(x))^2 = (2(sin(x)-cos(x)))/(sin(x)+cos(x))^3`

Explanation:

In this problem, we will use the following:
The chain rule:
`d/dxf(g(x)) = f'(g(x))g'(x)`

The power rule:
`d/dxx^n = nx^(n-1)`

and the following derivatives:
`d/dx sin(x) = cos(x)`
`d/dx cos(x) = -sin(x)`

To avoid needing the quotient rule, we first write
`d/dx1/(sin(x) + cos(x))^2 = d/dx(sin(x)+cos(x))^-2`

Next, by the chain rule and power rule, we have
`d/dx(sin(x)+cos(x))^-2 = -2(sin(x)+cos(x))^-3(d/dx(sin(x)+cos(x)))`

`=>d/dx(sin(x)+cos(x))^-2 =`
`-2(sin(x)+cos(x))^-3(cos(x)-sin(x))`

so, bringing the `-1` into the numerator, our final result is

`d/dx1/(sin(x) + cos(x))^2 = (2(sin(x)-cos(x)))/(sin(x)+cos(x))^3`