Differentiate ${x^x}$ from the first principle where $x > 0$.
A. ${x^x}[1 + \log x]$
B. ${x^x}$
C. ${x^{x - 1}}$
D. none of these
Answer
652.8k+ views
Hint: To solve this question, we will use the first principle method which is the most basic method to differentiate a given function. The formula to find derivatives using the first principle is $\mathop {\lim }\limits_{h \to 0} \dfrac{{f(a + h) - f(a)}}{h}$.
Now, for a real valued function f and a is a point in the domain of function. The derivative of function f is defined as,
$\mathop {\lim }\limits_{h \to 0} \dfrac{{f(a + h) - f(a)}}{h}$
Complete step-by-step answer:
Now, we have f = ${x^x}$. So, using the above formula, we get
$\dfrac{{df}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{(x + h)}^{x + h}} - {x^x}}}{h}$
Adding and subtracting ${x^{x + h}}$ in the above expression, we get
\[\dfrac{{df}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}({(x + h)^{x + h}} - {x^{x + h}} + {x^{x + h}} - {x^x})\]
\[\dfrac{{df}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}({x^{x + h}}\{ {(\dfrac{{x + h}}{x})^x}.{(\dfrac{{x + h}}{x})^h} - 1\} + {x^x}({x^h} - 1))\]
Taking ${x^x}$ common from the above expression, we get
\[\dfrac{{df}}{{dx}} = {x^x}\mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ {{{\left\{ {{{\left( {1 + \dfrac{h}{x}} \right)}^{\dfrac{x}{h}}}} \right\}}^h} - 1} \right] + Lt\left[ {\dfrac{{{x^h} - 1}}{h}} \right]\]
So, we get
\[\dfrac{{df}}{{dx}} = {x^x}\left[ {\lim \dfrac{{{e^h} - 1}}{h} + Lt\dfrac{{{x^h} - 1}}{h}} \right]\]
Now, using the property, $\lim \dfrac{{{a^x} - 1}}{x} = \log a$, the above expression can be written as,
$\dfrac{{df}}{{dx}} = {x^x}[\log e + \log x]$. Now, loge = 1,
Therefore, $\dfrac{{df}}{{dx}} = {x^x}[1 + \log x]$
So, option (A) is correct.
Note: Whenever we come up with such types of questions, we will have to use the formula $\mathop {\lim }\limits_{h \to 0} \dfrac{{f(a + h) - f(a)}}{h}$ to find the derivative of the given function using first principle. When we solve such types of questions, we have to follow a few steps. First, we will put the value of the function in the formula of first principle. After it, we will simplify the expression formed by adding and subtracting a term which helps us in taking a common from the expression. Like in this question, we add and subtract ${x^{x + h}}$ so that we can take ${x^x}$ common and thus our expression is simplified. After it, we will use some properties such as $\lim \dfrac{{{a^x} - 1}}{x} = \log a$ to solve the question.
Now, for a real valued function f and a is a point in the domain of function. The derivative of function f is defined as,
$\mathop {\lim }\limits_{h \to 0} \dfrac{{f(a + h) - f(a)}}{h}$
Complete step-by-step answer:
Now, we have f = ${x^x}$. So, using the above formula, we get
$\dfrac{{df}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{(x + h)}^{x + h}} - {x^x}}}{h}$
Adding and subtracting ${x^{x + h}}$ in the above expression, we get
\[\dfrac{{df}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}({(x + h)^{x + h}} - {x^{x + h}} + {x^{x + h}} - {x^x})\]
\[\dfrac{{df}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}({x^{x + h}}\{ {(\dfrac{{x + h}}{x})^x}.{(\dfrac{{x + h}}{x})^h} - 1\} + {x^x}({x^h} - 1))\]
Taking ${x^x}$ common from the above expression, we get
\[\dfrac{{df}}{{dx}} = {x^x}\mathop {\lim }\limits_{h \to 0} \dfrac{1}{h}\left[ {{{\left\{ {{{\left( {1 + \dfrac{h}{x}} \right)}^{\dfrac{x}{h}}}} \right\}}^h} - 1} \right] + Lt\left[ {\dfrac{{{x^h} - 1}}{h}} \right]\]
So, we get
\[\dfrac{{df}}{{dx}} = {x^x}\left[ {\lim \dfrac{{{e^h} - 1}}{h} + Lt\dfrac{{{x^h} - 1}}{h}} \right]\]
Now, using the property, $\lim \dfrac{{{a^x} - 1}}{x} = \log a$, the above expression can be written as,
$\dfrac{{df}}{{dx}} = {x^x}[\log e + \log x]$. Now, loge = 1,
Therefore, $\dfrac{{df}}{{dx}} = {x^x}[1 + \log x]$
So, option (A) is correct.
Note: Whenever we come up with such types of questions, we will have to use the formula $\mathop {\lim }\limits_{h \to 0} \dfrac{{f(a + h) - f(a)}}{h}$ to find the derivative of the given function using first principle. When we solve such types of questions, we have to follow a few steps. First, we will put the value of the function in the formula of first principle. After it, we will simplify the expression formed by adding and subtracting a term which helps us in taking a common from the expression. Like in this question, we add and subtract ${x^{x + h}}$ so that we can take ${x^x}$ common and thus our expression is simplified. After it, we will use some properties such as $\lim \dfrac{{{a^x} - 1}}{x} = \log a$ to solve the question.
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