Find $\frac{\text{d}y}{\text{d}x}$ for each of the following, simplifying your answers.

1. Apply the chain rule and the standard formula for the derivative of inverse sine.

$\frac{\text{d}y}{\text{d}x} = \frac{1}{\sqrt{1 - (e^x)^2}} \cdot \frac{\text{d}}{\text{d}x}(e^x)$

2. Simplify the expression.

$\frac{\text{d}y}{\text{d}x} = \frac{e^x}{\sqrt{1 - e^{2x}}}$

1. Simplify the expression using laws of logarithms to make differentiation easier.

$y = \ln(1+\ln x) - \ln(x^x)$
$y = \ln(1+\ln x) - x\ln x$

2. Differentiate the terms using the chain rule and product rule.

$\frac{\text{d}y}{\text{d}x} = \frac{1}{1+\ln x} \cdot \frac{\text{d}}{\text{d}x}(1+\ln x) - \left( x \cdot \frac{\text{d}}{\text{d}x}(\ln x) + \ln x \cdot \frac{\text{d}}{\text{d}x}(x) \right)$
$\frac{\text{d}y}{\text{d}x} = \frac{1}{x(1+\ln x)} - \left( x \left(\frac{1}{x}\right) + \ln x \right)$
$\frac{\text{d}y}{\text{d}x} = \frac{1}{x(1+\ln x)} - (1 + \ln x)$

3. Simplify the derivative into a single algebraic fraction.

$\frac{\text{d}y}{\text{d}x} = \frac{1 - x(1+\ln x)(1+\ln x)}{x(1+\ln x)}$
$\frac{\text{d}y}{\text{d}x} = \frac{1 - x(1+\ln x)^2}{x(1+\ln x)}$