Question

Find the value of $\dfrac{d}{{dx}}\left[ {{x^x} + {x^a} + {a^x} + {a^a}} \right]$.
$\left( 1 \right){x^x}\left( {1 + \log x} \right) + a.{x^{a - 1}} \\\ \left( 2 \right){x^x}\left( {1 + \log x} \right) + a.{x^{a - 1}} + {a^x}.\log a \\\ \left( 3 \right){x^x}\left( {1 + \log x} \right) + {a^a}(1 + \log a) \\\ \left( 4 \right){x^x}\left( {1 + \log x} \right) + {a^a}(1 + \log a) + a.{x^{a - 1}} + {a^a}(1 + \log a) \\\$

Answer 1

The power rule states that $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n.{x^{n - 1}}$.
The derivative of $\log x$ is given by $\dfrac{1}{x},x \ne 0$.
The derivative of a constant is always $0$.
The derivative of an exponential function for a variable is equal to the product of the exponential function and natural logarithm of the base of the exponential function.
$\therefore \dfrac{d}{{dx}}\left( {{a^x}} \right) = {a^x}.{\log _e}a$

Complete step-by-step answer:
It has asked to find the differentiation of the expression $({x^x} + {x^a} + {a^x} + {a^a})$.
So we have to first find the derivative of the individual terms and then combine them to get the derivative of the expression.
First, we find the derivative of the term ${x^x}$.
Let $y = {x^x}$.
Taking $\log$ on both sides of the equation we get:
$\log y = \log {x^x} \\\ \Rightarrow \log y = x\log x \\\$
Differentiating both sides of the expression with respect to $x$ we get:
$\dfrac{d}{{dx}}\left( {\log y} \right) = \dfrac{d}{{dx}}\left( {x\log x} \right) \\\ \Rightarrow \dfrac{1}{y} \times \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( x \right) \times \log x + \dfrac{d}{{dx}}\left( {\log x} \right) \times x \\\ \Rightarrow \dfrac{1}{y} \times \dfrac{{dy}}{{dx}} = 1 \times \log x + \dfrac{1}{x} \times x \\\ \Rightarrow \dfrac{{dy}}{{dx}} = y \times \left( {\log x + 1} \right) \\\ \Rightarrow \dfrac{{dy}}{{dx}} = {x^x}\left( {\log x + 1} \right) \\\$
Next using the power rule we get the derivative of ${x^a}$.
$\dfrac{d}{{dx}}\left( {{x^a}} \right) = a \times {x^{a - 1}}$
The derivative of the exponential function ${a^x}$ is given by ${a^x} \times {\log _e}a$ i.e. the exponential function multiplied by the natural logarithm of the base of the exponential function.
Lastly, ${a^a}$ is a constant. This is because a constant raised to the power of a constant will also be a constant value.
So, the derivative of constant is $0$.
Therefore, to find the differentiation of the required expression we sum up the derivatives of all the terms.
$\therefore$ The required differentiation is :
${x^x}\left( {\log x + 1} \right) + a \times {x^{a - 1}} + {a^x} \times \log a$.
Hence, the correct option is $\left( 2 \right){x^x}\left( {1 + \log x} \right) + a.{x^{a - 1}} + {a^x}.\log a$.

Note: One should be well acquainted with the derivative formulas to find the differentiation of a big expression. In the case of bigger expressions, we always find the derivatives of small terms and then concatenate them to get the differentiation of the big expression. One should not get confused with the ${a^a}$ term because if a constant is raised by any power then it will remain as a constant. So its derivative will also be $0$.