# What is the derivative of this function y=x^2+cos^-1x?

Feb 17, 2017

$y ' = 2 x - \frac{1}{\sin} x$

#### Explanation:

$y = {x}^{2} + {\cos}^{-} 1 x$
$y = {x}^{2} + \frac{1}{\cos} x$
$\frac{d}{\mathrm{dx}} y = \frac{d}{\mathrm{dx}} \left({x}^{2} + \frac{1}{\cos} x\right)$
$y ' = 2 x + \left(\frac{1}{-} \sin x\right)$
$y ' = 2 x - \frac{1}{\sin} x$

you have to remember the trig table of derivatives
$\frac{d}{\mathrm{dx}} \sin x = \cos x$
$\frac{d}{\mathrm{dx}} \cos x = - \sin x$