# What is the derivative of 2^arcsin(x)?

Aug 3, 2016

$\frac{{2}^{\arcsin} \left(x\right) \ln \left(2\right)}{\sqrt{1 - {x}^{2}}}$

#### Explanation:

Let:

$y = {2}^{\arcsin} \left(x\right)$

Take the natural logarithm of both sides:

$\ln \left(y\right) = \ln \left({2}^{\arcsin} \left(x\right)\right)$

Simplify using logarithm rules:

$\ln \left(y\right) = \arcsin \left(x\right) \cdot \ln \left(2\right)$

Differentiate both sides. You should remember that:

• The left-hand side will need the chain rule, similar to implicit differentiation.
• On the right-hand side, $\ln \left(2\right)$ is just a constant.
• The derivative of $\arcsin \left(x\right)$ is $\frac{1}{\sqrt{1 - {x}^{2}}}$.

Differentiating yields:

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

Now, solving for $\frac{\mathrm{dy}}{\mathrm{dx}}$, the derivative, multiply both sides by $y$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \cdot \ln \left(2\right)}{\sqrt{1 - {x}^{2}}}$

Write $y$ as ${2}^{\arcsin} \left(x\right)$:

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