Find the first derivative of \(y = x^x\) for \(x \gt 0 \) with all the steps presented.
Note that the function \( y = x^{x}\) is neither a power function of the form \( x^{k}\) nor an exponential function of the form \( b^{x}\) and the known formulas of
Differentiation of these two functions
cannot be used. We need to find another method to find the first derivative of the given function.
Given:
\[ y = x^{x} \]
Take the natural log (ln) of both sides of the above:
\[ \ln y = \ln\left(x^{x}\right) \]
Use properties of logarithmic functions \(\ln A^{b} = b \ln A\) to the right side of the above equation and obtain:
\[ \ln y = x \ln x \]
Differentiate both sides of the above with respect to \(x\), using the chain rule on the left side and the product rule on the right:
\[ \frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}(x \ln x) \]
\[ \frac{1}{y} \cdot \frac{dy}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} \]
Simplify the right side:
\[ \frac{1}{y} \cdot \frac{dy}{dx} = \ln x + 1 \]
Multiply both sides by \(y\) and simplify:
\[ \frac{dy}{dx} = y(\ln x + 1) \]
Substitute \(y\) by \(x^{x}\) to obtain the final answer:
\[
\boxed{\frac{dy}{dx} = x^{x}(\ln x + 1)}
\]
Find the first derivative of:
\[ f(x) = x^{2x} \]
\[ g(x) = (\sin x)^{x} \]
\[ h(x) = (x^2 + 1)^{x} \]
Answers to the Above Exercises:
\[ f'(x) = 2x^{2x}(\ln x + 1) \]
\[ g'(x) = (\sin x)^{x}\left(\ln(\sin x) + \frac{x \cos x}{\sin x}\right) \]
\[ h'(x) = (x^2 + 1)^{x}\left(\ln(x^2 + 1) + \frac{2x^2}{x^2 + 1}\right) \]