# How do you find the derivative of 1/cosx?

May 14, 2016

It depends on what tools (theorems and definitions) you have to work with.

#### Explanation:

If we need to use the definition of derivative, we'll need the fundamental trigonometric limits:

${\lim}_{h \rightarrow 0} \frac{\sinh}{h} = 1$ $\text{ }$ and $\text{ }$ ${\lim}_{h \rightarrow 0} \frac{1 - \cosh}{h} = 0$

along with ${\lim}_{h \rightarrow 0} \cos \left(x + h\right) = \cos x$.

We'll also need the trigonometric identity:

$\cos \left(x + h\right) = \cos x \cos h - \sin x \sin h$.

Here is the work.

For $f \left(x\right) = \frac{1}{\cos} x$, the definition of derivative gives

$f ' \left(x\right) = \frac{\frac{1}{\cos} \left(x + h\right) - \frac{1}{\cos} x}{h}$.

As always, the initial form of the limit of this difference quotient is indeterminate. We need to work with the difference quotient until we get a limit of determinate form.

We'll start by getting a single fractional expression.

$\frac{\frac{1}{\cos} \left(x + h\right) - \frac{1}{\cos} x}{h} = \frac{\left(\frac{1}{\cos} \left(x + h\right) - \frac{1}{\cos} x\right)}{h} \cdot \frac{\left(\cos \left(x + h\right) \cos x\right)}{\left(\cos \left(x + h\right) \cos x\right)}$

$= \frac{\cos x - \cos \left(x + h\right)}{h \cos \left(x + h\right) \cos x}$

Now expand $\cos \left(x + h\right)$ and distribute the subtraction.

$= \frac{\cos x - \cos x \cos h + \sin x \sin h}{h \cos \left(x + h\right) \cos x}$

Regroup so we can use thr fundamental trig limits.

$= \frac{\cos x \frac{1 - \cos h}{h} + \sin x \sin \frac{h}{h}}{\cos \left(x + h\right) \cos x}$

Evaluate the limit by evaluating the individual limits

${\lim}_{h \rightarrow 0} \frac{\cos x \frac{1 - \cos h}{h} + \sin x \sin \frac{h}{h}}{\cos \left(x + h\right) \cos x} = \frac{\cos x \left(0\right) + \sin x \left(1\right)}{\cos x \cos x}$

$= \sin \frac{x}{\cos} ^ 2 x = \frac{1}{\cos} x \sin \frac{x}{\cos} x = \sec x \tan x$

If you have the quotient rule

$\frac{d}{\mathrm{dx}} \left(\frac{1}{\cos} x\right) = \frac{0 \cos x - 1 \left(- \sin x\right)}{\cos x} ^ 2 = \sin \frac{x}{\cos} ^ 2 x$

Finish as above.

If you have the chain rule

Know that $\frac{d}{\mathrm{dx}} f {\left(x\right)}^{-} 1 = - f {\left(x\right)}^{-} 2 \cdot f ' \left(x\right)$, so

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

And again finish as above.

If you know the derivative of secant , use $\frac{1}{\cos} x = \sec x$