# What is the derivative of this function y=cos^-1(-2x^3-3)^3?

Oct 5, 2016

d/dx(cos^-1u(x))=(18x^2(-2x^3-3)^2)/(sqrt(1-(-2x^3-3)^6)

#### Explanation:

Based on the derivative on inverse trigonometric functions we have:

color(blue)(d/dx(cos^-1u(x))=-(d/dx(u(x)))/(sqrt(1-u(x)^2))
So, let us find $\frac{d}{\mathrm{dx}} \left(u \left(x\right)\right)$

Here ,$u \left(x\right)$ is a composite of two functions so we should apply chain rule to compute its derivative.

Let
$g \left(x\right) = - 2 {x}^{3} - 3$ and
$f \left(x\right) = {x}^{3}$

We have $u \left(x\right) = f \left(g \left(x\right)\right)$
The chain rule says:
color(red)(d/dx(u(x))=color(green)(f'(g(x)))*color(brown)(g'(x))

Let us find color(green)(f'(g(x))

$f ' \left(x\right) = 3 {x}^{2}$ then,
$f ' \left(g \left(x\right)\right) = 3 g {\left(x\right)}^{2}$
color(green)(f'(g(x))=3(-2x^3-3)^2

Let us find $\textcolor{b r o w n}{g ' \left(x\right)}$

$\textcolor{b r o w n}{g ' \left(x\right) = - 6 {x}^{2}}$

$\textcolor{red}{\frac{\mathrm{du} \left(x\right)}{\mathrm{dx}}} = \textcolor{g r e e n}{f ' \left(g \left(x\right)\right)} \cdot \textcolor{b r o w n}{g ' \left(x\right)}$

$\textcolor{red}{\frac{\mathrm{du} \left(x\right)}{\mathrm{dx}}} = \textcolor{g r e e n}{3 {\left(- 2 {x}^{3} - 3\right)}^{2}} \cdot \left(\textcolor{b r o w n}{- 6 {x}^{2}}\right)$
$\textcolor{red}{\frac{\mathrm{du} \left(x\right)}{\mathrm{dx}}} = - 18 {x}^{2} {\left(- 2 {x}^{3} - 3\right)}^{2}$

color(blue)(d/dx(cos^-1u(x))=-(d/dx(u(x)))/(sqrt(1-u(x)^2)

color(blue)(d/dx(cos^-1u(x))=-(-18x^2(-2x^3-3)^2)/(sqrt(1-((-2x^3-3)^3)^2)
Therefore,

color(blue)(d/dx(cos^-1u(x))=(18x^2(-2x^3-3)^2)/(sqrt(1-(-2x^3-3)^6)