What is the derivative of sec x?

Jun 5, 2016

It is $\sin \frac{x}{\cos} {\left(x\right)}^{2}$.

Explanation:

$\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$

So we want to calculate

$\frac{d}{\mathrm{dx}} \frac{1}{\cos} \left(x\right) = \frac{d}{\mathrm{dx}} \left(\cos {\left(x\right)}^{-} 1\right)$

for the chain rule this is equal to

$\frac{d}{\mathrm{dx}} \left(\cos {\left(x\right)}^{-} 1\right) = - \cos {\left(x\right)}^{-} 2 \cdot \frac{d}{\mathrm{dx}} \cos \left(x\right)$

$= - \frac{1}{\cos} {\left(x\right)}^{2} \cdot \left(- \sin \left(x\right)\right)$

$= \sin \frac{x}{\cos} {\left(x\right)}^{2}$

or, if you prefer, it is

$= \tan \left(x\right) \sec \left(x\right)$.

Jun 5, 2016

$\frac{d}{\mathrm{dx}} \sec x = \sec x \tan x$

Explanation:

To find the derivative of secant, we could either use the limit definition of the derivative (which would take a very long time) or the definition of secant itself:
$\sec x = \frac{1}{\cos} x$

We know $\frac{d}{\mathrm{dx}} \cos x = - \sin x$ - keep that in mind because we're going to need it.

Our problem is:
$\frac{d}{\mathrm{dx}} \sec x$

Since $\sec x = \frac{1}{\cos} x$, we can write this as:
$\frac{d}{\mathrm{dx}} \frac{1}{\cos} x$

We can find this derivative using the quotient rule:
$\frac{d}{\mathrm{dx}} \frac{u}{v} = \frac{u ' v - u v '}{v} ^ 2$

In our case, $u = 1 \to u ' = 0$ and $v = \cos x \to v ' = - \sin x$:
$\frac{d}{\mathrm{dx}} \frac{1}{\cos} x = \frac{\left(0\right) \left(\cos x\right) - \left(1\right) \left(- \sin x\right)}{\cos x} ^ 2$
$= \sin \frac{x}{\cos} ^ 2 x$

This is equivalent to:
$\frac{1}{\cos} x \cdot \sin \frac{x}{\cos} x$

Because $\frac{1}{\cos} x = \sec x$ and $\sin \frac{x}{\cos} x = \tan x$, this is:
$\sec x \tan x$

Therefore,
$\frac{d}{\mathrm{dx}} \sec x = \frac{d}{\mathrm{dx}} \frac{1}{\cos} x = \sec x \tan x$

Pro Tip
You can use this method to find the derivative of $\csc x$ and $\cot x$ by recognizing $\csc x = \frac{1}{\sin} x$ and $\cot x = \frac{1}{\tan} x$ and using the quotient rule, as we did above. This way, you only have to memorize the derivatives of sine, cosine, and tangent - you can derive the other three.