Question

How do you find the derivative of `y = ln(secx)`?

Answer 1

See below

Explanation:

In order to differentiate this function, we must use the chain rule:

`(f g(x))'=f'(g(x))xxg'(x)`.

Informally, this means that if we have to derivate a composite function, `fg(x)`, then we differentiate `f` with respect to `x`, treating `g(x)` as if it were `x` and then multiply that derivative by `g'(x)`.

The function to derivate is `lnsecx`. So `(lnsecx)'=ln'(secx)xx(secx)'=1/secx xxsecxtanx=tanx`

Answer 2

`dy/dx=tanx`

Explanation:

We can also rewrite this using logarithm rules:

`y=ln(secx)=ln(1/cosx)=ln((cosx)^-1)=-ln(cosx)`

The derivative of `ln(x)` is `1/x`, so according to the chain rule the derivative of `ln(f(x))` is `1/f(x)*f'(x)`.

Thus,

`dy/dx=-1/cosx*d/dxcosx`

The derivative of `cosx` is `-sinx`:

`dy/dx=-1/cosx(-sinx)=sinx/cosx=tanx`