# What is the derivative of f(x) = arcsin(2x^3 - 1)?

Apr 13, 2018

(df)/(dx)=(3x^2)/(xsqrt(x(1-x^3))

#### Explanation:

We can use here chain formula. As differential of $\arcsin x$ is $\frac{1}{\sqrt{1 - {x}^{2}}}$

derivative of $f \left(x\right) = \arcsin \left(2 {x}^{3} - 1\right)$ is

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {\left(2 {x}^{3} - 1\right)}^{2}}} \cdot \frac{d}{\mathrm{dx}} \left(2 {x}^{3} - 1\right)$

= $\frac{1}{\sqrt{1 - 4 {x}^{6} + 4 {x}^{3} - 1}} \cdot 6 {x}^{2}$

= (3x^2)/(xsqrt(x(1-x^3))

Apr 13, 2018

(3x^2)/sqrt(x^3(1-x^3)

#### Explanation:

We use the chain rule, which states that,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Let $u = 2 {x}^{3} - 1 , \therefore \frac{\mathrm{du}}{\mathrm{dx}} = 6 {x}^{2}$.

Then $y = \arcsin u , \implies \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{\sqrt{1 - {u}^{2}}}$.

Combining together, we get,

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

$= \frac{6 {x}^{2}}{\sqrt{1 - {u}^{2}}}$

Final step is to substitute back $u = 2 {x}^{3} - 1$, and we get,

$= \frac{6 {x}^{2}}{\sqrt{1 - {\left(2 {x}^{3} - 1\right)}^{2}}}$

=(3x^2)/sqrt(x^3(1-x^3)