# What is the derivative of  arcsin(x-1)?

Apr 12, 2018

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

#### Explanation:

derivative of inverse trigonometric functions

the general formula to differentiate the arcsin functions is

$\int {\sin}^{-} 1 u = \frac{1}{\sqrt{1 - {u}^{2}}} \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{d}{\mathrm{dx}} {\sin}^{-} 1 \left(x - 1\right) = \frac{1}{\sqrt{1 - {\left(x - 1\right)}^{2}}} \cdot \frac{d \left(x - 1\right)}{\mathrm{dx}}$ "rarr chain rule

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

Apr 13, 2018

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

#### Explanation:

We got:

$\frac{d}{\mathrm{dx}} \left(\arcsin \left(x - 1\right)\right)$

Let $y = \arcsin \left(x - 1\right)$

Let's use the chain rule, which states that,

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

Let $u = x - 1 , \therefore \frac{\mathrm{du}}{\mathrm{dx}} = 1$

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

Combining,

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

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

Substitute back $u = x - 1$ to get the final answer:

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