February 4, 2025 · 9:17 am

Differentiating the Tangent Function

Remember $tan(x)=\frac{sin(x)}{cos(x)}$ .

I use the quotient rule to differentiate $f(x)=tan(x)$ .

(1) $\begin{equation*}\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^2}\end{equation*}$

If $h(x)=tan(x)=\frac{sin(x)}{cos(x)}$ then from equation $1$

(2) $\begin{equation*}h'(x)=\frac{cos(x)cos(x)-(-sin(x)sin(x))}{[cos(x)]^2}\end{equation*}$

(3) $\begin{equation*}h'(x)=\frac{cos^2(x)+sin^2(x)}{cos^2(x)}\end{equation*}$

Remember the Pythagorean identity

(4) $\begin{equation*}sin^2(x)+cos^2(x)=1\end{equation*}$

Hence

$\begin{equation*}h'(x)=\frac{1}{cos^2(x)}=sec^2(x)\end{equation}$

(5) $\begin{equation*}\frac{d}{dx}tan(x)=sec^2(x)\end{equation*}$

Recent Posts