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# How do you differentiate $y={{x}^{x}}$?

Last updated date: 13th Jun 2024
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Hint: Take natural log $\log \left( {{\log }_{e}} \right)$ on both the sides and in the R.H.S. use the formula: - $\log {{a}^{m}}=m\log a$ to simplify. Now, consider the R.H.S. as the product of an algebraic function and a logarithmic function. Apply the product rule of differentiation given as: - $\dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$, where u = x and $v={{\log }_{e}}x$. In the L.H.S. use chain rule of differentiation given as: - $\dfrac{d\ln y}{dx}=\dfrac{d\ln y}{dy}\times \dfrac{dy}{dx}$ to simplify. Find the value of $\dfrac{dy}{dx}$ by using the formula: - $\dfrac{d\ln y}{dy}=\dfrac{1}{y}$ and substituting back the value $y={{x}^{x}}$.

Complete step-by-step solution:
Here, we have been provided with the function $y={{x}^{x}}$ and we are asked to differentiate it, that means we have to find the value of $\dfrac{dy}{dx}$.
We have $y={{x}^{x}}$, here in the R.H.S. We have an exponential form of the function where both the base and exponent are variable. So, taking natural log, i.e., log to the base e, both the sides, we get,
$\Rightarrow {{\log }_{e}}y={{\log }_{e}}\left( {{x}^{x}} \right)$
Using the formula: - $\log {{a}^{m}}=m\log a$, we get,
$\Rightarrow \ln y=x\ln x$
Now, we can assume the R.H.S. as the product of an algebraic function (x) and a logarithmic function ($\ln x$). So, we have,
$\Rightarrow \ln y=u\times v$
Differentiating both the sides with respect to x, we get,
$\Rightarrow \dfrac{d\ln y}{dx}=\dfrac{d\left( u\times v \right)}{dx}$
Now, applying the product rule of differentiation given as: - $\dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$, we get,
$\Rightarrow \dfrac{d\ln y}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$
Substituting the assumed values of u and v, we get,
$\Rightarrow \dfrac{d\ln y}{dx}=x\dfrac{d\ln x}{dx}+\ln x\dfrac{dx}{dx}$
We know that $\dfrac{d\ln x}{dx}=\dfrac{1}{x}$, so we have,
\begin{align} & \Rightarrow \dfrac{d\ln y}{dx}=x\times \dfrac{1}{x}+\ln x\times 1 \\ & \Rightarrow \dfrac{d\ln y}{dx}=1+\ln x \\ \end{align}
Using the chain rule of differentiation in the L.H.S., we have,
\begin{align} & \Rightarrow \dfrac{d\ln y}{dy}\times \dfrac{dy}{dx}=1+\ln x \\ & \Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\left( 1+\ln x \right) \\ \end{align}
Multiplying both the sides with y, we get,
$\Rightarrow \dfrac{dy}{dx}=y\left( 1+\ln x \right)$
Substituting back the value of y, we get,
$\Rightarrow \dfrac{dy}{dx}={{x}^{x}}\left( 1+\ln x \right)$
Hence, the derivative of the given function is ${{x}^{x}}\left( 1+\ln x \right)$, which is our answer.

Note: One may note that there can be a direct method to solve the question with taking logarithm both the sides. What we can do is we will differentiate ${{x}^{x}}$ by considering the exponent as a constant and use the formula: - $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$. Now, in the next step we will consider the base as a constant and use the formula: - $\dfrac{d{{a}^{x}}}{dx}={{a}^{x}}\log a$ to get the second part of the derivative. Finally, we will take the sum $n{{x}^{n-1}}+{{a}^{x}}\log a$ to get the answer, i.e., the value of $\dfrac{dy}{dx}$. You must remember all the basic rules of differentiation such as: - the product rule, the chain rule, $\dfrac{u}{v}$ rule etc.