If y = frac{A}{x} + B x^2, then x^2 frac{d^2 | vedclass

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

Question

(A) Given the equation: $y = A x^{-1} + B x^2$. First, differentiate $y$ with respect to $x$: $\frac{d y}{d x} = -A x^{-2} + 2 B x$. Next, differentia

Vedclass · Accepted Answer

$2 y$

Vedclass · Answer

(A) Given the equation: $y = A x^{-1} + B x^2$. First, differentiate $y$ with respect to $x$: $\frac{d y}{d x} = -A x^{-2} + 2 B x$. Next, differentiate again with respect to $x$ to find the second derivative: $\frac{d^2 y}{d x^2} = 2 A x^{-3} + 2 B$. Now, multiply by $x^2$: $x^2 \frac{d^2 y}{d x^2} = x^2 (2 A x^{-3} + 2 B) = 2 A x^{-1} + 2 B x^2$. Factor out $2$: $x^2 \frac{d^2 y}{d x^2} = 2 (A x^{-1} + B x^2)$. Since $y = A x^{-1} + B x^2$, we substitute $y$ back into the expression: $x^2 \frac{d^2 y}{d x^2} = 2 y$.

If $y = \frac{A}{x} + B x^2$, then $x^2 \frac{d^2 y}{d x^2} =$

Similar Questions

If $\cos ^{-1}\left(\frac{y}{b}\right)=2 \log \left(\frac{x}{2}\right)$,where $x>0$,then $x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}$ is equal to

If $f(x)=\tan ^{-1}\left[\frac{\log \left(\frac{e}{x^{2}}\right)}{\log \left(e x^{2}\right)}\right]+\tan ^{-1}\left[\frac{3+2 \log x}{1-6 \log x}\right]$ then the value of $f^{\prime \prime}(x)$ is equal to

If $f(x)=1+nx+\frac{n(n-1)}{2}x^2+\frac{n(n-1)(n-2)}{6}x^3+\ldots+x^n$,then $f''(1)$ is equal to

If $y = \sin x + e^x,$ then $\frac{d^2x}{dy^2} = $

If $y = \frac{\sinh^{-1} x}{\sqrt{1+x^2}}$,then $(1+x^2) y_2 + 3xy_1 + y = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module