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

Hint: 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.