To differentiate the function $y = x^{\tan^{-1} x}$ with respect to $x$, we'll need to use logarithmic differentiation since the exponent is a function of $x$.

Step-by-Step Solution

Given:

$y = x^{\tan^{-1} x}$

1. Take the natural logarithm of both sides:

$\ln y = \ln \left( x^{\tan^{-1} x} \right)$

2. Use the logarithm power rule:

$\ln y = \tan^{-1} x \cdot \ln x$

3. Differentiate both sides with respect to $x$:

$\frac{d}{dx} (\ln y) = \frac{d}{dx} \left( \tan^{-1} x \cdot \ln x \right)$

The derivative of $\ln y$ with respect to $x$ is:

$\frac{1}{y} \cdot \frac{dy}{dx}$

Now, apply the product rule to differentiate the right-hand side:

$\frac{1}{y} \cdot \frac{dy}{dx} = \left( \frac{d}{dx} (\tan^{-1} x) \right) \cdot \ln x + \left( \frac{d}{dx} (\ln x) \right) \cdot \tan^{-1} x$

Calculate the derivatives:

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

4. Solve for $\frac{dy}{dx}$:

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

Substitute back the value of $y = x^{\tan^{-1} x}$:

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

Final Answer

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

Would you like further details or have any questions?

Related Questions:

1. How do you differentiate functions involving inverse trigonometric functions?
2. What is logarithmic differentiation and when is it useful?
3. How does the product rule apply in calculus?
4. What are the common derivatives of inverse trigonometric functions?
5. How do you handle the differentiation of expressions with variables in both the base and the exponent?

Tip:

Remember that logarithmic differentiation is particularly useful when differentiating functions where both the base and the exponent involve variables.