Question

Find value of \[\dfrac{d}{{dx}}\left( {{x^x} + {x^a} + {a^x} + {a^a}} \right)\]
A. \[{x^x}\left( {1 + log{\text{ }}x} \right) + a{x^{a - 1}}\]
B. \[{x^x}\left( {1 + log{\text{ }}x} \right) + a{x^{a - 1}} + {a^x}\log {\text{ }}a\]
C. \[{x^x}\left( {1 + log{\text{ }}x} \right) + {a^a}\left( {1 + \log {\text{ a}}} \right)\]
D. \[{x^x}\left( {1 + log{\text{ }}x} \right) + {a^a}\left( {1 + \log {\text{ a}}} \right) + a{x^{a - 1}} + {a^a}\left( {1 + \log {\text{ a}}} \right)\]

Answer 1

Hint: In this question, we need to differentiate the given function with respect to \[x\]. For finding the derivative, we have to assume certain terms for simplification and take a log for easy differentiation. Here, we also have to use product rules for differentiation.

Formula Used:
$\dfrac{d x^a}{dx} = ax^{a-1}$
$\dfrac{d a^x}{dx} = x log x$

Complete answer:
 Let \[y = \left( {{x^x} + {x^a} + {a^x} + {a^a}} \right)\]
Now, let us differentiate \[y\]with respect to \[x\].
Thus, we get
\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^x}} \right) + \dfrac{d}{{dx}}\left( {{x^a}} \right) + \dfrac{d}{{dx}}\left( {{a^x}} \right) + \dfrac{d}{{dx}}\left( {{a^a}} \right)\]
We know that the derivative of any constant term is always zero.
Thus, we get
\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^x}} \right) + \dfrac{d}{{dx}}\left( {{x^a}} \right) + \dfrac{d}{{dx}}\left( {{a^x}} \right) + 0\]
Let us assume that \[u = {x^x}\]
Now, take log on both sides.
\[\log u = x\left( {\log x} \right)\]
Now, differentiate with respect to \[x\].
\[\dfrac{1}{u}\dfrac{{du}}{{dx}} = x\left( {\dfrac{1}{x}} \right) + \log x\]
\[ \Rightarrow \dfrac{{du}}{{dx}} = u\left( {1 + \log x} \right)\]
Now, put again \[u = {x^x}\]in the above equation.
\[ \Rightarrow \dfrac{{du}}{{dx}} = {x^x}\left( {1 + \log x} \right)\]
Finally, we get
\[\dfrac{{dy}}{{dx}} = {x^x}\left( {1 + \log x} \right) + a{x^{a - 1}} + {a^x}\left( {\log a} \right)\]
Hence, the result of \[\dfrac{d}{{dx}}\left( {{x^x} + {x^a} + {a^x} + {a^a}} \right)\]is \[{x^x}\left( {1 + \log x} \right) + a{x^{a - 1}} + {a^x}\left( {\log a} \right)\].
Therefore, the option (B) is correct.

Additional Information: Differentiation is a technique of determining the derivative of a function. That means, here we can find the instant rate of change in function based on one of its variables. Logarithmic differentiation refers to the process of differentiating functions by first obtaining logarithms and afterward differentiating. This method provides for the efficient calculation of derivatives of power, rational, and certain irrational functions. That means, logarithmic derivatives are used to make functions simpler before differentiation.


Note: Many students make mistakes in finding the derivative of \[{x^x}\]. Also, it is necessary to use proper chain rule or product rule to find the derivative of a function. The logarithmic differentiation plays a significant role in this case.