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

Suppose for a differentiable function $h$,$h(0)=0$,$h(1)=1$ and $h^{\prime}(0)=h^{\prime}(1)=2$. If $g(x)=h(e^{x}) e^{h(x)}$,then $g^{\prime}(0)$ is equal to:

Find the derivative of the following function (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\frac{ax+b}{px^{2}+qx+r}$

Find the derivative of the following function: $\frac{\sec x-1}{\sec x+1}$

If $f(x) = \log_{x^2} (\ln x)$,then $f'(e)$ is equal to . . . . . .

Find the derivative of the following function (it is to be understood that $p, q, r$ are fixed non-zero constants): $\frac{p x^{2}+q x+r}{x}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module