# How do you find the derivative of y=arcsin(5x+5)?

Mar 26, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5}{\sqrt{1 - {\left(5 x + 5\right)}^{2}}}$

#### Explanation:

If $u = \arcsin v$ then $v = \sin u$, hence $\frac{\mathrm{dv}}{\mathrm{du}} = \cos u$ and

$\frac{\mathrm{du}}{\mathrm{dv}} = \frac{1}{\cos} u = \frac{1}{\sqrt{1 - {\sin}^{2} u}} = \frac{1}{\sqrt{1 - {v}^{2}}}$

Hence, we can now find the the derivative of $y = \arcsin \left(5 x + 5\right)$ using the concept function of a function and this is given by

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {\left(5 x + 5\right)}^{2}}} \times \frac{d}{\mathrm{dx}} \left(5 x + 5\right)$ or

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5}{\sqrt{1 - {\left(5 x + 5\right)}^{2}}}$