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Question: If \[y = {({x^x})^x}\], then \[\dfrac{{dy}}{{dx}} = \] A) \[{({x^x})^x}(1 + 2\log x)\] B) \[{({x...

If y=(xx)xy = {({x^x})^x}, then dydx=\dfrac{{dy}}{{dx}} =
A) (xx)x(1+2log⁡x){({x^x})^x}(1 + 2\log x)
B) (xx)x(1−2log⁡x){({x^x})^x}(1 - 2\log x)
C) x(xx)x(1+2log⁡x)x{({x^x})^x}(1 + 2\log x)
D) x(xx)x(1−2log⁡x)x{({x^x})^x}(1 - 2\log x)

Explanation

Solution

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

Formula used: (am)n=am+n{({a^m})^n} = {a^{m + n}}
log⁡am=mlog⁡a\log {a^m} = m\log a
ddx[f(x)g(x)]=f(x)g′(x)+f′(x)g(x)\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=(xx)xy = {({x^x})^x}
We know that, (am)n=am+n{({a^m})^n} = {a^{m + n}}
Using the formula, we get,
y=xx2y = {x^{{x^2}}}
We know that; log⁡am=mlog⁡a\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(xx)x(1+2log⁡x)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)f(x) and g(x)g(x) be two functions of x.x. so, we can write that;
ddx[f(x)g(x)]=f(x)g′(x)+f′(x)g(x)\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⁡am=mlog⁡a\log {a^m} = m\log a