frac{d}{d x}(x^{2 x}) = . . . . . . ,x 0 | vedclass

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Question

(A) Let $y = x^{2 x}$.Taking the natural logarithm on both sides,we get $\log y = \log(x^{2 x})$.Using the property $\log(a^b) = b \log a$,we have $\l

Vedclass · Accepted Answer

$2 x^{2 x}(1+\log x)$

Vedclass · Answer

(A) Let $y = x^{2 x}$.Taking the natural logarithm on both sides,we get $\log y = \log(x^{2 x})$.Using the property $\log(a^b) = b \log a$,we have $\log y = 2 x \log x$.Differentiating both sides with respect to $x$ using the product rule:$\frac{1}{y} \frac{d y}{d x} = 2 \cdot \frac{d}{d x}(x \log x)$.$\frac{1}{y} \frac{d y}{d x} = 2 \left( x \cdot \frac{1}{x} + \log x \cdot 1 \right)$.$\frac{1}{y} \frac{d y}{d x} = 2(1 + \log x)$.Therefore,$\frac{d y}{d x} = y \cdot 2(1 + \log x)$.Substituting $y = x^{2 x}$,we get $\frac{d y}{d x} = 2 x^{2 x}(1 + \log x)$.Thus,the correct option is $A$.

$\frac{d}{d x}(x^{2 x}) =$ . . . . . . ,$x > 0$

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