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Question

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

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

Explanation

Solution

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.vu.v rule and d(1og(y))dx=1y.dydx\,\dfrac{d(1og(y))}{dx}\,=\dfrac{1}{y}\,\,.\dfrac{dy}{dx} we can simplify the answer.
After simplifying you will get the value of dydx\dfrac{dy}{dx}\,\,\,\, .

Complete step by step answer:
Here, according to the question
y=xx(1)y={{x}^{x}}----(1)
Take the log on both side on equation (1)(1) we get:
log(y)=log(xx)(2)\log (y)=\log ({{x}^{x}})---(2)
By using the property of log(xx)=xlog(x)(3)\log ({{x}^{x}})=x\log (x)--(3)
By substituting the value of equation (3)(3) in equation (2)(2)
log(y)=xlog(x)(4)\log (y)=x\log (x)---(4)
Take the derivative on both side
d(1og(y))dx=d(xlog(x))dx\dfrac{d(1og(y))}{dx}\,=\dfrac{d(x\log (x))}{dx}
d(xlog(x))dx=x×1x+1×log(x)\dfrac{d(x\log (x))}{dx}=x\times \dfrac{1}{x}+1 \times \log (x) using product rule.
Substituting the value of d(1og(y))dx=1y.dydx\,\dfrac{d(1og(y))}{dx}\,=\dfrac{1}{y}\,\,.\dfrac{dy}{dx} and d(xlog(x))dx=1+log(x)\dfrac{d(x\log (x))}{dx}=1+\log (x) in equation (4)(4)
1y.dydx=1+log(x)\dfrac{1}{y}\,\,.\dfrac{dy}{dx}=1+\log (x)
After further simplifying we get:
dydx=y(1+log(x))(5)\dfrac{dy}{dx}=y(1+\log (x))----(5)
After substituting the value of equation (1)(1) in equation (5)(5)
dydx=xx(1+log(x))\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)\sin (x) or any other (except power of x) we can take the derivative directly and solve the problem.