If y^2 = ax^2 + bx + c,where a, b, c are cons | vedclass

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

Question

(C) Given $y^2 = ax^2 + bx + c$. Differentiating with respect to $x$,we get: $2y \frac{dy}{dx} = 2ax + b$ $\frac{dy}{dx} = \frac{2ax + b}{2y}$ Differe

Vedclass · Accepted Answer

a constant

Vedclass · Answer

(C) Given $y^2 = ax^2 + bx + c$. Differentiating with respect to $x$,we get: $2y \frac{dy}{dx} = 2ax + b$ $\frac{dy}{dx} = \frac{2ax + b}{2y}$ Differentiating again with respect to $x$ using the quotient rule: $\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{2ax + b}{2y} \right) = \frac{2a(2y) - (2ax + b)(2 \frac{dy}{dx})}{4y^2}$ Substituting $\frac{dy}{dx} = \frac{2ax + b}{2y}$: $\frac{d^2y}{dx^2} = \frac{4ay - (2ax + b) \frac{2ax + b}{y}}{4y^2} = \frac{4ay^2 - (2ax + b)^2}{4y^3}$ Now,multiply by $y^3$: $y^3 \frac{d^2y}{dx^2} = \frac{4ay^2 - (2ax + b)^2}{4} = \frac{4a(ax^2 + bx + c) - (4a^2x^2 + 4abx + b^2)}{4}$ $y^3 \frac{d^2y}{dx^2} = \frac{4a^2x^2 + 4abx + 4ac - 4a^2x^2 - 4abx - b^2}{4} = \frac{4ac - b^2}{4}$ Since $a, b, c$ are constants,$\frac{4ac - b^2}{4}$ is a constant.

If $y^2 = ax^2 + bx + c$,where $a, b, c$ are constants,then $y^3 \frac{d^2 y}{dx^2}$ is equal to

Similar Questions

If $e^y + xy = e$,then the value of $\frac{d^2y}{dx^2}$ for $x = 0$ is

Variables $x$ and $y$ are related by the equation $x = \int\limits_0^y \frac{dt}{\sqrt{1 + t^2}}$. The value of $\frac{d^2y}{dx^2}$ is equal to

$A$ function $f$ satisfies the relation $f(x) = f''(x) + f'''(x) + \dots \infty$,where $f(x)$ is an infinitely differentiable function. If $f(1) = 5$,then the value of $f'(1) + f''(1)$ is equal to:

If $f(x)=x^3-x^2 f^{\prime}(1)+x f^{\prime \prime}(2)-f^{\prime \prime \prime}(3), x \in R$,then

$\frac{d^2}{dx^2}(2\cos x \cos 3x) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module