# How do you differentiate y=2cos(x)+6cos^-1(x)?

Jun 17, 2017

Given: $y = 2 \cos \left(x\right) + 6 {\cos}^{-} 1 \left(x\right)$

Differentiate each term:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d \left(2 \cos \left(x\right)\right)}{\mathrm{dx}} + \frac{d \left(6 {\cos}^{-} 1 \left(x\right)\right)}{\mathrm{dx}}$

Bring the constants outside:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \frac{d \left(\cos \left(x\right)\right)}{\mathrm{dx}} + 6 \frac{d \left({\cos}^{-} 1 \left(x\right)\right)}{\mathrm{dx}}$

We know that $\frac{d}{\mathrm{dx}} \cos \left(x\right) = - \sin \left(x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 \sin \left(x\right) + 6 \frac{d \left({\cos}^{-} 1 \left(x\right)\right)}{\mathrm{dx}}$

We know that $\frac{d}{\mathrm{dx}} {\cos}^{-} 1 \left(x\right) = - \frac{1}{\sqrt{1 - {x}^{2}}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 \sin \left(x\right) - \frac{6}{\sqrt{1 - {x}^{2}}}$

Jun 17, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 \sin x - \frac{6}{\sqrt{1 - {x}^{2}}}$

#### Explanation:

$\text{using the "color(blue)"standard derivatives}$

• d/dx(cosx)=-sinx

• d/dx(cos^-1x)=-1/(sqrt(1-x^2)

$y = 2 \cos x + 6 {\cos}^{-} 1 x$

rArrdy/dx=-2sinx-6/(sqrt(1-x^2)