Question

How do you find the derivative of `y = secx/(1 + tanx)`?

Answer 1

`dy/dx = (sinx - cosx)/(sinx + cosx)^2`

Explanation:

First of all, I'm assuming the function is `y = secx/(1 + tanx)`? Call your function `f(x)`.

`f(x) = secx/(1 + tanx)`

`f(x) = (1/cosx)/(1 + sinx/cosx)`

`f(x) = (1/cosx)/((cosx + sinx)/cosx)`

`f(x) = 1/cosx xx cosx/(cosx + sinx)`

`f(x) = 1/(cosx+ sinx)`

`f(x) = (cosx + sinx)^-1`

We let `y = u^-1` and `u = cosx + sinx`. Then `dy/(du) = -1u^(-2)` and `(du)/dx = -sinx + cosx`.

The chain rule states that `color(red)(dy/dx = dy/(du) xx (du)/dx`.

`dy/dx = -1/u^2 xx -sinx + cosx`

`dy/dx = (sinx - cosx)/(sinx + cosx)^2`

Hopefully this helps!