# Question 59763

Jan 12, 2018

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

#### Explanation:

To start:

${\lim}_{\Delta x \to 0} \frac{\sec \left(x + \Delta x\right) - \sec \left(x\right)}{\Delta x}$

$= {\lim}_{\Delta x \to 0} \frac{\frac{1}{\cos} \left(x + \Delta x\right) - \frac{1}{\cos} \left(x\right)}{\Delta x}$

$= {\lim}_{\Delta x \to 0} \frac{\frac{\cos \left(x\right) - \cos \left(x + \Delta x\right)}{\cos \left(x\right) \cos \left(x + \Delta x\right)}}{\Delta x}$

$= {\lim}_{\Delta x \to 0} \frac{\cos \left(x\right) - \cos \left(x + \Delta x\right)}{\cos \left(x\right) \cos \left(x + \Delta x\right) \Delta x}$

Now use the trig identity:

$\cos \left(A\right) - \cos \left(B\right) = - 2 \sin \left(\frac{A + B}{2}\right) \sin \left(\frac{A - B}{2}\right)$

So we now have:

=lim_(Deltax->0)(-2sin((x+x+Deltax)/2)sin((x - x-Deltax)/2))/(cos(x)cos(x+Deltax)Deltax)#

$= {\lim}_{\Delta x \to 0} \frac{- 2 \sin \left(x + \frac{\Delta x}{2}\right) \sin \left(- \frac{\Delta x}{2}\right)}{\cos \left(x\right) \cos \left(x + \Delta x\right) \Delta x}$

Take the negative at the front and cancel with the negative inside the second sin function and multiply the top and the bottom by $\frac{1}{2}$ to take the factor of $2$ at the top and place on the bottom like so:

$= {\lim}_{\Delta x \to 0} \frac{\sin \left(x + \frac{\Delta x}{2}\right) \sin \left(\frac{\Delta x}{2}\right)}{\cos \left(x\right) \cos \left(x + \Delta x\right) \frac{\Delta x}{2}}$

$= {\lim}_{\Delta x \to 0} \frac{\sin \left(x + \frac{\Delta x}{2}\right)}{\cos \left(x\right) \cos \left(x + \Delta x\right)} \cdot {\lim}_{\Delta x \to 0} \sin \frac{\frac{\Delta x}{2}}{\frac{\Delta x}{2}}$

${\lim}_{\Delta x \to 0} \sin \frac{\frac{\Delta x}{2}}{\frac{\Delta x}{2}} = 1$

So we are left with:

$= {\lim}_{\Delta x \to 0} \frac{\sin \left(x + \frac{\Delta x}{2}\right)}{\cos \left(x\right) \cos \left(x + \Delta x\right)}$

Evaluating the limit by direct substitution:

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

That is the derivative of $\sec \left(x\right)$ so that is exactly what we expected.

For a proof that:

${\lim}_{\Delta x \to 0} \sin \frac{\frac{\Delta x}{2}}{\frac{\Delta x}{2}} = 1$

Jan 12, 2018

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

#### Explanation:

To evaluate ${\lim}_{\Delta x \to 0} \left(\frac{\sec \left(x + \Delta x\right) - \sec \left(x\right)}{\Delta x}\right)$ notice that is of the form ${\lim}_{\Delta x \to 0} \left(\frac{f \left(x + \Delta x\right) - f \left(x\right)}{\Delta x}\right)$, which is the limit definition of the derivative.

So,

${\lim}_{\Delta x \to 0} \left(\frac{\sec \left(x + \Delta x\right) - \sec \left(x\right)}{\Delta x}\right) = f ' \left(x\right)$,

where $f \left(x\right) = \sec \left(x\right)$.

For $f \left(x\right)$ it's known that $f ' \left(x\right) = \sec \left(x\right) \tan \left(x\right)$ so:

${\lim}_{\Delta x \to 0} \left(\frac{\sec \left(x + \Delta x\right) - \sec \left(x\right)}{\Delta x}\right) = \sec \left(x\right) \tan \left(x\right)$.

Jan 12, 2018

Very much like Andrews I.'s answer. The details are different.

#### Explanation:

We'll need:
$\sec t = \frac{1}{\cos} t$ and

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

and the limits:

${\lim}_{\Delta x \rightarrow 0} \frac{\sin \Delta x}{\Delta x} = 1$ and ${\lim}_{\Delta x \rightarrow 0} \frac{1 - \cos \Delta x}{\Delta x} = 0$.

Solution

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

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

Expand $\cos \left(x + \Delta x\right)$ in the numerator only. (Its would be OK to expand it in the denominator, but it is not necessary.)

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

Regroup to use the fundamental trigonometric limits.

$= {\lim}_{\Delta x \rightarrow 0} \left[\frac{\cos x - \cos x \cos \Delta x}{\Delta x \cos x \cos \left(x + \Delta x\right)} + \frac{\sin x \sin \Delta x}{\Delta x \cos \left(x\right) \cos \left(x + \Delta x\right)}\right]$

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

Evaluate the limit using the fundamental trig limits and continuity of cosine.

$= \cos \frac{x}{\cos x \cos \left(x\right)} \left(0\right) + \frac{\sin x}{\cos \left(x\right) \cos \left(x\right)} \left(1\right)$

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