If y = 2x^{3x},then frac{dy}{dx} at x = 1 is | vedclass

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(B) Given $y = 2x^{3x}$.Taking natural logarithm on both sides: $\ln y = \ln(2x^{3x}) = \ln 2 + 3x \ln x$.Differentiating both sides with respect to $

Vedclass · Accepted Answer

$6$

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(B) Given $y = 2x^{3x}$.Taking natural logarithm on both sides: $\ln y = \ln(2x^{3x}) = \ln 2 + 3x \ln x$.Differentiating both sides with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = 0 + 3 \ln x + 3x \cdot \frac{1}{x} = 3 \ln x + 3$.Thus,$\frac{dy}{dx} = y(3 \ln x + 3) = 2x^{3x}(3 \ln x + 3)$.At $x = 1$,$\frac{dy}{dx} = 2(1)^{3(1)}(3 \ln 1 + 3)$.Since $\ln 1 = 0$,we have $\frac{dy}{dx} = 2(1)(0 + 3) = 2 	imes 3 = 6$.

If $y = 2x^{3x}$,then $\frac{dy}{dx}$ at $x = 1$ is

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