Question

Find the derivation of ${x^x} + {a^x} + {x^a} + {a^a}$ for some fixed $x > 0$, $a > 0$.

Answer 1

Derivation of expression is the rate of change of a function with respect to independent variables. Derivation of a function is basically a measure of sensitivity to change of function value with change in the argument where argument refers to the input whose output is to be found. Derivatives are useful in finding the slope of an equation, maxima, and minima of a function when the slope is zero and is also used to check a function, whether it is increasing or decreasing.
For example, $y = 3{x^2} + 2x + 1$hence the differentiation of y with respect to $x$ will be:

$$\dfrac{{dy}}{{dx}} = 3 \times 2 \times {x^{2 - 1}} + 2{x^{1 - 1}} + 0 \\\ = 6{x^1} + 2{x^0} \\\ = 6x + 2 \\\$$

Formula used: Basic formulae of derivation to be used in this question are:

$$\dfrac{d}{{dx}}\left( c \right) = 0 \\\ \dfrac{d}{{dx}}\left( x \right) = 1 \\\ \dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} \\\ \dfrac{{d\left( {{a^x}} \right)}}{{dx}} = {a^{^x}}\ln a \\\$$

Complete step by step answer:
We are considering $x$ to be the variable part of the equation and $a$ to be the constant part.
Let $y = {x^x} + {a^x} + {x^a} + {a^a}$ now differentiate it with respect to $x$, we get
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^x}} \right) + {a^x}\ln a + a{x^{a - 1}} + 0$--- (i), where $a$ is a constant term
Now let us assume $u = {x^x}$, and if we take log on both the sides we get:
$\ln u = x\ln x$--- (ii)
Now differentiate the equation (ii) with respect to $x$, we get

$$\dfrac{d}{{dx}}\left( {\ln u} \right) = \dfrac{d}{{dx}}\left( {x\ln x} \right) \\\ \\\$$

To solve the differentiation of $x\ln x$, use the chain rule, which is $\dfrac{d}{{dx}}\left( {ab} \right) = a\dfrac{d}{{dx}}\left( b \right) + b\dfrac{d}{{dx}}\left( a \right)$. Hence we get:

$$\dfrac{1}{u}\dfrac{{du}}{{dx}} = x \times \dfrac{{d\left( {\ln x} \right)}}{{dx}} + \left( {\ln x} \right)\dfrac{d}{{dx}}\left( x \right) \\\ \dfrac{1}{u}\dfrac{{du}}{{dx}} = x \times \dfrac{1}{x} + \ln x \\\ \dfrac{1}{u}\dfrac{{du}}{{dx}} = 1 + \ln x \\\$$

Now substitute the value of $u = {x^x}$ we get in the above solved equation:

$$\dfrac{{du}}{{dx}} = u\left( {1 + \ln x} \right) \\\ \dfrac{{d{x^x}}}{{dx}} = {x^x}\left( {1 + \ln x} \right) \\\$$

Now put $\dfrac{{d{x^x}}}{{dx}} = {x^x}\left( {1 + \ln x} \right)$ in equation (i), we will have:
$\dfrac{{dy}}{{dx}} = {x^x}\left( {1 + \ln x} \right) + {a^x}\ln a + a{x^{a - 1}}$
Where$x > 0$,$a > 0$.

Note: As a constant term does not contain any variables with them when they are differentiated, then their value is zero. Derivation of a function is represented in$\dfrac{a}{b}$, where $a$ is the function which is being differentiated and b its independent variable by which function is being differentiated written as $\dfrac{{dy}}{{dx}}$ where y is the function.