We are tasked with finding the derivative of the equation:

$y^x + x^{\cos(y)} = \arcsin\left(\frac{2x}{1 + x^2}\right)$

Step 1: Implicit differentiation

To differentiate implicitly, we apply the derivative to both sides with respect to $x$, treating $y$ as a function of $x$, i.e., $y = y(x)$.

Differentiate the left-hand side:

1. First term: $y^x$

   • We apply logarithmic differentiation: $\frac{d}{dx}\left(y^x\right) = y^x \left( \ln(y) + \frac{x}{y} \frac{dy}{dx} \right)$

2. Second term: $x^{\cos(y)}$

   • Again, apply logarithmic differentiation: $\frac{d}{dx}\left(x^{\cos(y)}\right) = x^{\cos(y)} \left( \frac{\cos(y)}{x} - \sin(y) \frac{dy}{dx} \ln(x) \right)$

Differentiate the right-hand side:

The derivative of the right-hand side is: $\frac{d}{dx}\left( \arcsin\left(\frac{2x}{1 + x^2}\right) \right)$ Let $z = \frac{2x}{1 + x^2}$, then: $\frac{d}{dx}\left(\arcsin(z)\right) = \frac{1}{\sqrt{1 - z^2}} \cdot \frac{d}{dx}(z)$ Now, differentiate $z$: $\frac{d}{dx} \left(\frac{2x}{1 + x^2}\right) = \frac{(1 + x^2) \cdot 2 - 2x \cdot 2x}{(1 + x^2)^2} = \frac{2(1 - x^2)}{(1 + x^2)^2}$ Thus: $\frac{d}{dx} \arcsin\left(\frac{2x}{1 + x^2}\right) = \frac{1}{\sqrt{1 - \left( \frac{2x}{1 + x^2} \right)^2}} \cdot \frac{2(1 - x^2)}{(1 + x^2)^2}$ Since $1 - \left( \frac{2x}{1 + x^2} \right)^2 = \frac{(1 + x^2)^2 - 4x^2}{(1 + x^2)^2} = \frac{(1 - x^2)^2}{(1 + x^2)^2}$, the derivative becomes: $\frac{2(1 - x^2)}{(1 + x^2) \cdot (1 - x^2)} = \frac{2}{1 + x^2}$

Step 2: Combine the derivatives

Now, combine everything and solve for $\frac{dy}{dx}$:

$y^x \left( \ln(y) + \frac{x}{y} \frac{dy}{dx} \right) + x^{\cos(y)} \left( \frac{\cos(y)}{x} - \sin(y) \frac{dy}{dx} \ln(x) \right) = \frac{2}{1 + x^2}$

Group the terms containing $\frac{dy}{dx}$ and solve:

$\frac{dy}{dx} \left( y^x \frac{x}{y} - x^{\cos(y)} \sin(y) \ln(x) \right) = \frac{2}{1 + x^2} - y^x \ln(y) - x^{\cos(y)} \frac{\cos(y)}{x}$

Thus, the derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = \frac{\frac{2}{1 + x^2} - y^x \ln(y) - x^{\cos(y)} \frac{\cos(y)}{x}}{y^x \frac{x}{y} - x^{\cos(y)} \sin(y) \ln(x)}$

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Let me know if you want further clarification or details.

Related Questions:

1. How would you solve $y^x = k$ for $y$ explicitly?
2. What happens when $y$ is constant, and how does that simplify the derivative?
3. Can you differentiate other functions like $e^{x^y}$ implicitly?
4. How would you apply the chain rule for $f(g(x)) = \arcsin(x)$?
5. Could this problem be solved using logarithmic differentiation for both sides?

Tip: Always group terms involving $\frac{dy}{dx}$ when performing implicit differentiation to simplify the solving process.