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

Apr 1, 2017

See below

#### Explanation:

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

$\left(f g \left(x\right)\right) ' = f ' \left(g \left(x\right)\right) \times g ' \left(x\right)$.

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

The function to derivate is $\ln \sec x$. So $\left(\ln \sec x\right) ' = \ln ' \left(\sec x\right) \times \left(\sec x\right) ' = \frac{1}{\sec} x \times \sec x \tan x = \tan x$

Apr 1, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \tan x$

#### Explanation:

We can also rewrite this using logarithm rules:

$y = \ln \left(\sec x\right) = \ln \left(\frac{1}{\cos} x\right) = \ln \left({\left(\cos x\right)}^{-} 1\right) = - \ln \left(\cos x\right)$

The derivative of $\ln \left(x\right)$ is $\frac{1}{x}$, so according to the chain rule the derivative of $\ln \left(f \left(x\right)\right)$ is $\frac{1}{f} \left(x\right) \cdot f ' \left(x\right)$.

Thus,

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\cos} x \cdot \frac{d}{\mathrm{dx}} \cos x$

The derivative of $\cos x$ is $- \sin x$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\cos} x \left(- \sin x\right) = \sin \frac{x}{\cos} x = \tan x$