# Find the derivative of f(x)=-tan(x)?

Feb 12, 2018

$f ' \left(x\right) = - {\sec}^{2} \left(x\right)$

#### Explanation:

We want to find the derivative of

$f \left(x\right) = - \tan \left(x\right)$

Use the definition $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$

$f \left(x\right) = - \sin \frac{x}{\cos} \left(x\right)$

Use the quotient rule, if $f \left(x\right) = \frac{h \left(x\right)}{g} \left(x\right)$,

then $f ' \left(x\right) = \frac{h ' \left(x\right) g \left(x\right) - h \left(x\right) g ' \left(x\right)}{h \left(x\right)} ^ 2$

By the quotient rule with $h \left(x\right) = \sin \left(x\right)$ and $g \left(x\right) = \cos \left(x\right)$

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

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

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

$= - \frac{1}{\cos} ^ 2 \left(x\right)$

$= - {\sec}^{2} \left(x\right)$