Question

If \[y = {({x^x})^x}\], then \[\dfrac{{dy}}{{dx}} = \]
A) \[{({x^x})^x}(1 + 2\log x)\]
B) \[{({x^x})^x}(1 - 2\log x)\]
C) \[x{({x^x})^x}(1 + 2\log x)\]
D) \[x{({x^x})^x}(1 - 2\log x)\]

Answer 1

Hint: To find the first derivative of ordinary differential means we have to find out the derivative of \[y\] with respect \[x\].
The process of determining the derivative of a function is known as differentiation. Here, we try to reduce the power of the function of \[x\] by using few formulas.
Here we will apply the formula of derivative of the multiplication of any two functions of \[x\].

Formula used: \[{({a^m})^n} = {a^{m + n}}\]
\[\log {a^m} = m\log a\]
\[\dfrac{d}{{dx}}[f(x)g(x)] = f(x)g'(x) + f'(x)g(x)\]

Complete step-by-step answer:
It is given that; \[y = {({x^x})^x}\]
We know that, \[{({a^m})^n} = {a^{m + n}}\]
Using the formula, we get,
\[y = {x^{{x^2}}}\]
We know that; \[\log {a^m} = m\log a\]
Taking log in both sides we get,
$\Rightarrow$\[\log y = {x^2}\log x\]
Let us consider \[f(x)\] and \[g(x)\] be two functions of \[x.\] so, we can write that;
$\Rightarrow$\[\dfrac{d}{{dx}}[f(x)g(x)] = f(x)g'(x) + f'(x)g(x)\]
Differentiating with respect to \[x\] we get,
$\Rightarrow$\[\dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{{{x^2}}}{x} + 2x\log x\]
Simplifying we get,
$\Rightarrow$\[\dfrac{1}{y}\dfrac{{dy}}{{dx}} = x + 2x\log x\]
Simplifying again we get,
$\Rightarrow$\[\dfrac{1}{y}\dfrac{{dy}}{{dx}} = x + 2x\log x\]
Simplifying again we get,
$\Rightarrow$\[\dfrac{{dy}}{{dx}} = y(x + 2x\log x)\]
Substitute the value of \[y = {({x^x})^x}\]we get,
$\Rightarrow$\[\dfrac{{dy}}{{dx}} = {({x^x})^x}(x + 2x\log x)\]
Simplifying again we get,
$\Rightarrow$\[\dfrac{{dy}}{{dx}} = x{({x^x})^x}(1 + 2\log x)\]

Hence, the correct option is C) \[x{({x^x})^x}(1 + 2\log x)\]

Note: The derivative of a function in calculus of variable standards the sensitivity to change the output value with respect to a change in its input value. Derivatives are a primary tool of calculus.
For example, the derivative of a moving object position as per time-interval is the object’s velocity. It measures the quick change of position of object or person as the time changes.
Let us consider \[f(x)\] and \[g(x)\] be two functions of \[x.\] so, we can write that;
\[\dfrac{d}{{dx}}[f(x)g(x)] = f(x)g'(x) + f'(x)g(x)\]
Logarithms are the opposite phenomena of exponential like subtraction is the inverse of addition process, and division is the opposite phenomena of multiplication. Logs “undo” exponentials.
We know that; \[\log {a^m} = m\log a\]