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How do you find the derivative of x2x?

Answer
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Hint: In this question, we have to find the derivative of the given quantity. Consider the given quantity equal to some variable and then take logarithm both the sides of the equation and use the property, logab=bloga. After that, differentiate it with respect to x using the product rule of the derivative. After that do simplification and substitute the value of y to get the desired result.

Complete step-by-step answer:
Let y=x2x ….. (1)
Take log on both sides we have,
logy=logx2x
Now according to logarithmic property logab=bloga, we can write the above equation as
logy=2xlogx
Now apply the chain rule of differentiation we have,
ddx(ab)=addx(b)+bddx(a)
So, differentiate the above equation with respect to x what we have,
ddx(logy)=2xddx(logx)+(logx)ddx(2x)
Differentiate the terms,
1ydydx=2xx+(logx)×2
Cancel out the common factor,
1ydydx=2+2logx
Take 2 commons on the right side,
1ydydx=2(1+logx)
Multiplying both sides by y, we get
dydx=2y(1+logx)
Substitute the value of y from equation (1),
dydx=2x2x(1+logx)

Hence, the derivative of x2x is 2x2x(1+logx).

Note:
Whenever we face such types of problems the key concept is simply to make use of logarithm, as exponential powers need to be changed to a simpler form before starting doing the derivative part. A good understanding of some logarithm properties, product rule of derivatives along properties of logarithm helps to get on the right track to reach the answer.
Logarithms can make multiplication and division of large numbers easier, because adding logarithms is the same as multiplying, and subtracting logarithms is the same as dividing.
Change of base rule law,
logyx=logxlogy
Product rule law,
logxy=logx+logy
Quotient rule law,
logxy=logxlogy
Power rule law,
logxy=ylogx