Find $\frac{d y}{d x}$, if $y^{x}+x^{y}+x^{x}=a^{b}$.

Step 1: Decompose the equation

The given equation involves a sum of three complex terms, each of which is a function raised to another function. To simplify the differentiation process, we can break down the equation into individual terms. Let $u = y^x$, $v = x^y$, and $w = x^x$. The equation then becomes $u+v+w = a^b$.

Step 2: Differentiate each term with respect to x

Since $a^b$ is a constant (as $a$ and $b$ are constants), its derivative with respect to $x$ is $0$. Therefore, we need to find the derivative of each term $u$, $v$, and $w$ with respect to $x$ and sum them up to $0$.

Step 3: Calculate $\frac{du}{dx}$

To differentiate $u = y^x$, we take the natural logarithm of both sides: $\ln u = x \ln y$. Then, we differentiate implicitly with respect to $x$. Using the product rule on the right side, we get $\frac{1}{u} \frac{du}{dx} = \ln y + x \frac{1}{y} \frac{dy}{dx}$. Finally, we multiply by $u$ to solve for $\frac{du}{dx}$.

Step 4: Calculate $\frac{dv}{dx}$

Similarly, for $v = x^y$, we take the natural logarithm: $\ln v = y \ln x$. Differentiating implicitly with respect to $x$ gives $\frac{1}{v} \frac{dv}{dx} = \frac{dy}{dx} \ln x + y \frac{1}{x}$. Multiplying by $v$ yields the expression for $\frac{dv}{dx}$.

Step 5: Calculate $\frac{dw}{dx}$

For $w = x^x$, we take the natural logarithm: $\ln w = x \ln x$. Differentiating implicitly with respect to $x$ gives $\frac{1}{w} \frac{dw}{dx} = \ln x + x \frac{1}{x}$. Multiplying by $w$ yields the expression for $\frac{dw}{dx}$.

Step 6: Substitute and solve for $\frac{dy}{dx}$

Now, substitute the expressions for $\frac{du}{dx}$, $\frac{dv}{dx}$, and $\frac{dw}{dx}$ back into the equation $\frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} = 0$. Group the terms containing $\frac{dy}{dx}$ on one side and the remaining terms on the other side. Finally, divide to solve for $\frac{dy}{dx}$.

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