Question

If $y={{x}^{x}}$ , then $\,\dfrac{dy}{dx}$ is
A. $\,{{x}^{x}}\log (ex)\,$
B. ${{x}^{x}}\left( 1+\dfrac{1}{x} \right)$
C. ${{x}^{x}}(1+\log (x))$
D. ${{x}^{x}}\log (x)$

Answer 1

In this particular problem we have to take log on both sides then take the derivative on both side and applying the derivative rule that $u.v$ rule and $\,\dfrac{d(1og(y))}{dx}\,=\dfrac{1}{y}\,\,.\dfrac{dy}{dx}$ we can simplify the answer.
After simplifying you will get the value of $\dfrac{dy}{dx}\,\,\,\,$ .

Complete step by step answer:
Here, according to the question
$y={{x}^{x}}----(1)$
Take the log on both side on equation $(1)$ we get:
$\log (y)=\log ({{x}^{x}})---(2)$
By using the property of $\log ({{x}^{x}})=x\log (x)--(3)$
By substituting the value of equation $(3)$ in equation $(2)$
$\log (y)=x\log (x)---(4)$
Take the derivative on both side
$\dfrac{d(1og(y))}{dx}\,=\dfrac{d(x\log (x))}{dx}$
$\dfrac{d(x\log (x))}{dx}=x\times \dfrac{1}{x}+1 \times \log (x)$ using product rule.
Substituting the value of $\,\dfrac{d(1og(y))}{dx}\,=\dfrac{1}{y}\,\,.\dfrac{dy}{dx}$ and $\dfrac{d(x\log (x))}{dx}=1+\log (x)$ in equation $(4)$
$\dfrac{1}{y}\,\,.\dfrac{dy}{dx}=1+\log (x)$
After further simplifying we get:
$\dfrac{dy}{dx}=y(1+\log (x))----(5)$
After substituting the value of equation $(1)$ in equation $(5)$
$\dfrac{dy}{dx}={{x}^{x}}(1+\log (x))$

So, the correct answer is “Option C”.

Note: In this particular sum we cannot take direct derivative of the question because here there is a power of x. Therefore, to solve such a type of problem we can apply log on both sides and by using some rules of derivative we can solve it easily. If the question is asked like $\sin (x)$ or any other (except power of x) we can take the derivative directly and solve the problem.