# What is the equation of the tangent line of f(x)=e^xsecx at x=pi/4?

Apr 15, 2018

$y = {e}^{\frac{\pi}{4}} \sqrt{2} \left(2 x + 1 - \frac{\pi}{2}\right)$

#### Explanation:

The slope of the tangent line to a function $f \left(x\right)$ at $x = {x}_{1}$ is given by $f ' \left({x}_{1}\right)$. Recall that the derivative of a function returns its slope at any given point.

We will calculate $f ' \left(x\right)$. See that we use the product rule.

$f ' \left(x\right) = {e}^{x} \sec x \tan x + {e}^{x} \sec x$
$f ' \left(x\right) = {e}^{x} \sec x \left(\tan x + 1\right)$

Evaluating $f ' \left(x\right)$ at $x = \frac{\pi}{4}$ gives

$f ' \left(\frac{\pi}{4}\right) = {e}^{\frac{\pi}{4}} \sec \left(\frac{\pi}{4}\right) \left(\tan \left(\frac{\pi}{4}\right) + 1\right) = {e}^{\frac{\pi}{4}} \left(\sqrt{2}\right) \left(1 + 1\right)$
$= {e}^{\frac{\pi}{4}} \left(\sqrt{2}\right) \left(2\right) = {e}^{\frac{\pi}{4}} 2 \sqrt{2}$

The slope of our tangent line, then, is $m = {e}^{\frac{\pi}{4}} 2 \sqrt{2}$. We wish to find the line tangent to the point of $f \left(x\right)$ where $x = \frac{\pi}{4}$. This point is given by $\left(\frac{\pi}{4} , f \left(\frac{\pi}{4}\right)\right) = \left(\frac{\pi}{4} , {e}^{\frac{\pi}{4}} \sqrt{2}\right)$.

Now that we have a slope and a point, we can use slope-intercept form to get our equation.

$y = m x + b$
$y = {e}^{\frac{\pi}{4}} 2 \sqrt{2} \left(x\right) + b$
${e}^{\frac{\pi}{4}} \sqrt{2} = {e}^{\frac{\pi}{4}} 2 \sqrt{2} \left(\frac{\pi}{4}\right) + b$
$b = {e}^{\frac{\pi}{4}} \sqrt{2} - {e}^{\frac{\pi}{4}} 2 \sqrt{2} \left(\frac{\pi}{4}\right)$
$b = {e}^{\frac{\pi}{4}} \sqrt{2} \left(1 - \frac{2 \pi}{4}\right) = {e}^{\frac{\pi}{4}} \sqrt{2} \left(1 - \frac{\pi}{2}\right)$

Having found $b$, we have all the ingredients of our linear equation.

$y = {e}^{\frac{\pi}{4}} 2 \sqrt{2} \left(x\right) + {e}^{\frac{\pi}{4}} \sqrt{2} \left(1 - \frac{\pi}{2}\right)$
$y = {e}^{\frac{\pi}{4}} \sqrt{2} \left(2 x + 1 - \frac{\pi}{2}\right)$