Find the derivative of y ^ x + x ^ (cos y) = arcsin((2x)/(1 + x ^ 2))

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.

Discover how to solve the derivative of the equation y^x + x^(cos(y)) = arcsin((2x)/(1 + x^2)) using implicit and logarithmic differentiation. Learn key calculus concepts such as the chain rule and inverse trigonometric differentiation. Suitable for undergraduate calculus students.

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