The Derivatives of Trigonometric Functions - Exercise 1


Exercise 1. Find the equations of the tangent line and the normal line to the graph of $f(x) = \sec(x) + \tan(x)$ at the point $\left(\displaystyle \frac{\pi}{4},
f\left(\frac{\pi}{4}\right)\right)$.

Answer. First we need to find the derivative of f(x) so we can get the slope of the tangent line and the normal line. We have

\begin{displaymath}f'(x) = \sec(x) \tan(x) + \sec^2(x)\;.\end{displaymath}

So we have

\begin{displaymath}f'\left(\frac{\pi}{4}\right) = \sqrt{2} + 2\end{displaymath}

knowing that $\sec(\pi/4) = \sqrt{2}$ and $\tan(\pi/4) = 1$. Note that

\begin{displaymath}f\left(\frac{\pi}{4}\right) = \sqrt{2} + 1\;.\end{displaymath}

So the equation of the tangent line at the point $(\pi/4,
\sqrt{2} + 1)$ is

\begin{displaymath}y - \sqrt{2} - 1 = (\sqrt{2} + 2) \left(x- \frac{\pi}{4}\right) \;.\end{displaymath}

the slope of the normal line to the graph at the point $(\pi/4,
\sqrt{2} + 1)$ is

\begin{displaymath}- \frac{1}{\sqrt{2} + 2}\end{displaymath}

which helps us find the equation of the normal line as

\begin{displaymath}y - \sqrt{2} - 1 = - \frac{1}{\sqrt{2} + 2} \left(x- \frac{\pi}{4}\right) \;.\end{displaymath}


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