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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$-6$

Vedclass · Answer

(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 $k =$

Similar Questions

Find the derivative of the following function (it is to be understood that $a, b, c,$ and $d$ are fixed non-zero constants): $\frac{a x+b}{c x+d}$

If $f(x) = \frac{x}{1+|x|}$,for $x \in \mathbb{R}$,then $f^{\prime}(0)$ is equal to :

If $y = x + \frac{1}{x}$,then which among the following holds?

$\frac{d}{d x}\left[\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2\right]=$ . . . . . .

$\frac{d}{dx} \log_{\sqrt{x}} \left(\frac{1}{x}\right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module