# Question a4762

Mar 12, 2017

$\frac{d}{\mathrm{dx}} \left(\tan x\right) = {\sec}^{2} x$

#### Explanation:

Using the $\textcolor{b l u e}{\text{trigonometric identities}}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\tan x = \frac{\sin x}{\cos x}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

differentiate using the $\textcolor{b l u e}{\text{quotient rule}}$

$\text{Given " f(x)=(g(x))/(h(x))" then}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{here } g \left(x\right) = \sin x \Rightarrow g ' \left(x\right) = \cos x$

$\text{and } h \left(x\right) = \cos x \Rightarrow h ' \left(x\right) = - \sin x$

$\Rightarrow f ' \left(x\right) = \frac{\cos x \left(\cos x\right) - \sin x \left(- \sin x\right)}{{\cos}^{2} x}$

$\textcolor{w h i t e}{\Rightarrow f ' \left(x\right)} = \frac{{\cos}^{2} x + {\sin}^{2} x}{{\cos}^{2} x}$

• cos^2x+sin^2x=1" and " 1/(cos^2x)=sec^2x#

$\Rightarrow \frac{d}{\mathrm{dx}} \left(\tan x\right) = \frac{1}{{\cos}^{2} x} = {\sec}^{2} x$

This result is a $\textcolor{b l u e}{\text{standard derivative}}$ and worth remembering.

$\textcolor{b l u e}{\text{Note:}}$ This is a Calculus question not Algebra