If y = log_{e} x^3 + 3 sin^{-1} x + k x^2 and | vedclass

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(B) Given $y = \log_{e} x^3 + 3 \sin^{-1} x + k x^2$. Using the property of logarithms,$y = 3 \log_{e} x + 3 \sin^{-1} x + k x^2$. Differentiating wit

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$-6$

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(B) Given $y = \log_{e} x^3 + 3 \sin^{-1} x + k x^2$. Using the property of logarithms,$y = 3 \log_{e} x + 3 \sin^{-1} x + k x^2$. Differentiating with respect to $x$,we get $y' = \frac{3}{x} + \frac{3}{\sqrt{1 - x^2}} + 2kx$. Given $y'(\frac{1}{2}) = 2 \sqrt{3}$,substitute $x = \frac{1}{2}$ into the derivative: $y'(\frac{1}{2}) = \frac{3}{1/2} + \frac{3}{\sqrt{1 - (1/2)^2}} + 2k(\frac{1}{2}) = 2 \sqrt{3}$. $6 + \frac{3}{\sqrt{3/4}} + k = 2 \sqrt{3}$. $6 + \frac{3}{\sqrt{3}/2} + k = 2 \sqrt{3}$. $6 + \frac{6}{\sqrt{3}} + k = 2 \sqrt{3}$. $6 + 2 \sqrt{3} + k = 2 \sqrt{3}$. $k = -6$.

If $y = \log_{e} x^3 + 3 \sin^{-1} x + k x^2$ and $y'(\frac{1}{2}) = 2 \sqrt{3}$,then $k =$

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