Math Doubts

Evaluate $\dfrac{d}{dx}{\big(x^{\displaystyle x}\big)}$ by first principle of differentiation

The derivative of x raised to the power of x with respect to x is evaluated by using the logarithms. It can also be calculated as per the fundamental definition of the derivatives. The differentiation of $x$-th power of $x$ can be expressed in limit form as per its first principle.

$\dfrac{d}{dx}{\big(x^{\displaystyle x}\big)}$ $\,=\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{(x+h)^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$

Separate the common factor from the term

Look at the exponential expression in the numerator, the first term is a binomial $x+h$ raised to the power of $x+h$ and the second term is $x$ to the $x$-th power.

$\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{(x+h)^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$

The first term can be expanded by using Binomial Theorem but it consists the expression $x$-th power of $x$ internally. So, let us try to separate it from the first term.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{(x \times 1 + 1 \times h)^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$

In order to separate the $x$ to the $x$-th power, the second term in the binomial should have a factor in terms of $x$. For this reason, the number $1$ can be written as a quotient of $x$ by $x$.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\bigg(x \times 1 + \dfrac{x}{x} \times h\bigg)^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\bigg(x \times 1 + \dfrac{x \times 1}{x} \times h\bigg)^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\bigg(x \times 1 + x \times \dfrac{1}{x} \times h\bigg)^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\bigg(x \times 1 + x \times \dfrac{1 \times h}{x}\bigg)^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\bigg(x \times 1 + x \times \dfrac{h}{x}\bigg)^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$

In the binomial of the first term, both terms consist of $x$ as a common factor. So, it can be taken out common from them.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\bigg(x \times \bigg(1+\dfrac{h}{x}\bigg)\bigg)^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$

The first term represents power of a product rule and use this formula to convert the first term as a product of two factors.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x+h} \times \bigg(1+\dfrac{h}{x}\bigg)^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$

Expand the function in exponential notation

The second factor in the first term can be expanded as per the Binomial Theorem in one variable.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x+h} \times \Bigg(1+(x+h) \times \bigg(\dfrac{h}{x}\bigg)+\dfrac{(x+h)(x+h-1)}{2!} \times \bigg(\dfrac{h}{x}\bigg)^2+\cdots\Bigg)-x^{\displaystyle x}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x+h} \times \Bigg(1+(x+h)\bigg(\dfrac{h}{x}\bigg)+\dfrac{(x+h)(x+h-1)}{2!}\bigg(\dfrac{h}{x}\bigg)^2+\cdots\Bigg)-x^{\displaystyle x}}{h}}$

Prepare the expression for finding the Limit

In the first term of the numerator, the $x$ raised to the power $x+h$ multiplies sum of infinite terms. So, it can be distributed to all the terms as per the distributive property of multiplication over the addition.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x+h} \times 1+x^{\displaystyle x+h} \times (x+h)\bigg(\dfrac{h}{x}\bigg)+x^{\displaystyle x+h} \times \dfrac{(x+h)(x+h-1)}{2!}\bigg(\dfrac{h}{x}\bigg)^2+\cdots-x^{\displaystyle x}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x+h}+x^{\displaystyle x+h} \times (x+h)\bigg(\dfrac{h}{x}\bigg)+x^{\displaystyle x+h} \times \dfrac{(x+h)(x+h-1)}{2!}\bigg(\dfrac{h}{x}\bigg)^2+\cdots-x^{\displaystyle x}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x+h}-x^{\displaystyle x}+x^{\displaystyle x+h} \times (x+h)\bigg(\dfrac{h}{x}\bigg)+x^{\displaystyle x+h} \times \dfrac{(x+h)(x+h-1)}{2!}\bigg(\dfrac{h}{x}\bigg)^2+\cdots}{h}}$

For our convenience, write the expression in the numerator as a sum of two expressions.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\Big(x^{\displaystyle x+h}-x^{\displaystyle x}\Big)+\Bigg(x^{\displaystyle x+h} \times (x+h)\bigg(\dfrac{h}{x}\bigg)+x^{\displaystyle x+h} \times \dfrac{(x+h)(x+h-1)}{2!}\bigg(\dfrac{h}{x}\bigg)^2+\cdots\Bigg)}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg(\dfrac{x^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$ $+$ $\dfrac{x^{\displaystyle x+h} \times (x+h)\bigg(\dfrac{h}{x}\bigg)+x^{\displaystyle x+h} \times \dfrac{(x+h)(x+h-1)}{2!}\bigg(\dfrac{h}{x}\bigg)^2+\cdots}{h}\Bigg)$

It is time to use the addition rule of limits to find the limit of sum of two terms.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$ $+$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x+h} \times (x+h)\bigg(\dfrac{h}{x}\bigg)+x^{\displaystyle x+h} \times \dfrac{(x+h)(x+h-1)}{2!}\bigg(\dfrac{h}{x}\bigg)^2+\cdots}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$ $+$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg(\dfrac{x^{\displaystyle x+h} \times (x+h)\bigg(\dfrac{h}{x}\bigg)}{h}+\dfrac{x^{\displaystyle x+h} \times \dfrac{(x+h)(x+h-1)}{2!}\bigg(\dfrac{h}{x}\bigg)^2}{h}+\cdots\Bigg)}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$ $+$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg(\dfrac{x^{\displaystyle x+h} \times (x+h)\bigg(\dfrac{1 \times h}{x}\bigg)}{h}+\dfrac{x^{\displaystyle x+h} \times \dfrac{(x+h)(x+h-1)}{2!}\bigg(\dfrac{1 \times h}{x}\bigg)^2}{h}+\cdots\Bigg)}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$ $+$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg(\dfrac{x^{\displaystyle x+h} \times (x+h)\bigg(\dfrac{1}{x}\bigg) \times h}{h}+\dfrac{x^{\displaystyle x+h} \times \dfrac{(x+h)(x+h-1)}{2!}\bigg(\dfrac{1}{x}\bigg)^2 \times h^2}{h}+\cdots\Bigg)}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$ $+$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg(\dfrac{x^{\displaystyle x+h} \times (x+h)\bigg(\dfrac{1}{x}\bigg) \times \cancel{h}}{\cancel{h}}+\dfrac{x^{\displaystyle x+h} \times \dfrac{(x+h)(x+h-1)}{2!}\bigg(\dfrac{1}{x}\bigg)^2 \times \cancel{h^2}}{\cancel{h}}+\cdots\Bigg)}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$ $+$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg(x^{\displaystyle x+h} \times (x+h)\bigg(\dfrac{1}{x}\bigg)+\dfrac{x^{\displaystyle x+h} \times (x+h)(x+h-1)}{2!}\bigg(\dfrac{1}{x}\bigg)^2 \times h+\cdots\Bigg)}$

Evaluate the Limit of each function

The limit of a rational function is successfully split as a sum of limits of two functions. Now, let’s find the limit of the infinite series firstly as the value of $h$ approaches $0$ by using the direct substitution method.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$ $+$ $x^{\displaystyle x+0} \times (x+0)\bigg(\dfrac{1}{x}\bigg)$ $+$ $\dfrac{x^{\displaystyle x+0} \times (x+0)(x+0-1)}{2!}\bigg(\dfrac{1}{x}\bigg)^2 \times 0$ $+$ $\cdots$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$ $+$ $x^{\displaystyle x} \times (x)\bigg(\dfrac{1}{x}\bigg)$ $+$ $\dfrac{x^{\displaystyle x} \times (x)(x-1)}{2!}\bigg(\dfrac{1}{x}\bigg)^2 \times 0$ $+$ $\cdots$

Due to the involvement of factor $h$ in each term of infinite series, all terms become zero except first term.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$ $+$ $x^{\displaystyle x} \times \bigg(\dfrac{x \times 1}{x}\bigg)$ $+$ $0$ $+$ $\cdots$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$ $+$ $x^{\displaystyle x} \times \bigg(\dfrac{x}{x}\bigg)$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$ $+$ $x^{\displaystyle x} \times \bigg(\dfrac{\cancel{x}}{\cancel{x}}\bigg)$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$ $+$ $x^{\displaystyle x} \times 1$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$ $+$ $x^{\displaystyle x}$

$=\,\,\,$ $x^{\displaystyle x}$ $+$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$

It is time to find the limit of remining function. The $x$ raised to the power of $x+h$ can be split as a product of two factors as per the product rule of exponents.

$=\,\,\,$ $x^{\displaystyle x}$ $+$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x} \times x^{\displaystyle h} -x^{\displaystyle x}}{h}}$

In both terms of the numerator, $x$-th power of $x$ is a common factor. So, take the common factor out from the terms for simplifying the numerator further.

$=\,\,\,$ $x^{\displaystyle x}$ $+$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x} \times x^{\displaystyle h} -x^{\displaystyle x} \times 1}{h}}$

$=\,\,\,$ $x^{\displaystyle x}$ $+$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle x} \times \big(x^{\displaystyle h} -1\big)}{h}}$

$=\,\,\,$ $x^{\displaystyle x}$ $+$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg(x^{\displaystyle x} \times \dfrac{x^{\displaystyle h} -1}{h}\Bigg)}$

According to the constant multiple rule of limits, the $x$ to the $x$-th power is a constant. Hence, it can be taken out from the limit operation.

$=\,\,\,$ $x^{\displaystyle x}$ $+$ $x^{\displaystyle x} \times \displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{x^{\displaystyle h} -1}{h}}$

According to the limit rule of a to the xth power minus 1 by x, the limit of $x$ raised to the power of $h$ minus $1$ by $h$ as value of $h$ approaches $0$ is equal to natural logarithm of $x$.

$=\,\,\,$ $x^{\displaystyle x}$ $+$ $x^{\displaystyle x} \times \log_{e}{x}$

The $x$-th power of $x$ is a common factor in both terms of the expression. So, take it out common from them for expressing the limit in simple form.

$=\,\,\,$ $x^{\displaystyle x} \times 1$ $+$ $x^{\displaystyle x} \times \log_{e}{x}$

$=\,\,\,$ $x^{\displaystyle x} \times \big(1+\log_{e}{x}\big)$

$=\,\,\,$ $x^{\displaystyle x}\big(1+\log_{e}{x}\big)$

$\therefore\,\,\,$ $\dfrac{d}{dx}{\big(x^{\displaystyle x}\big)}$ $\,=\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{(x+h)^{\displaystyle x+h}-x^{\displaystyle x}}{h}}$ $\,=\,$ $x^{\displaystyle x}\big(1+\log_{e}{x}\big)$

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