If y = x log left( frac{x}{a + bx} right),the | vedclass

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

Question

(B) Given $y = x \log \left( \frac{x}{a + bx} \right)$.Dividing by $x$,we get $\frac{y}{x} = \log x - \log(a + bx)$.Differentiating with respect to $x

Vedclass · Accepted Answer

$\left( x \frac{dy}{dx} - y \right)^2$

Vedclass · Answer

(B) Given $y = x \log \left( \frac{x}{a + bx} \right)$.Dividing by $x$,we get $\frac{y}{x} = \log x - \log(a + bx)$.Differentiating with respect to $x$:$\frac{x \frac{dy}{dx} - y}{x^2} = \frac{1}{x} - \frac{b}{a + bx} = \frac{a + bx - bx}{x(a + bx)} = \frac{a}{x(a + bx)}$.Thus,$x \frac{dy}{dx} - y = \frac{ax}{a + bx}$ ..... $(i)$.Differentiating equation $(i)$ with respect to $x$:$x \frac{d^2y}{dx^2} + \frac{dy}{dx} - \frac{dy}{dx} = \frac{(a + bx)a - ax(b)}{(a + bx)^2}$.$x \frac{d^2y}{dx^2} = \frac{a^2}{(a + bx)^2}$.Multiplying both sides by $x^2$:$x^3 \frac{d^2y}{dx^2} = \frac{a^2 x^2}{(a + bx)^2} = \left( \frac{ax}{a + bx} \right)^2$.From equation $(i)$,$\frac{ax}{a + bx} = x \frac{dy}{dx} - y$.Therefore,$x^3 \frac{d^2y}{dx^2} = \left( x \frac{dy}{dx} - y \right)^2$.

If $y = x \log \left( \frac{x}{a + bx} \right)$,then $x^3 \frac{d^2y}{dx^2} = $

Similar Questions

If $x^3 + y^3 = 6$ and $\frac{d^2y}{dx^2} \cdot \frac{d^2x}{dy^2} = \frac{m}{(xy)^n}$, where $m, n \in R$, then find the value of $\frac{m}{n}$.

If $y=\frac{2}{\sqrt{a^2-b^2}} \tan ^{-1}\left[\sqrt{\frac{a-b}{a+b}} \tan \frac{x}{2}\right]$,then $\left.\frac{d^2 y}{d x^2}\right|_{x=\frac{\pi}{2}}=$

If $y=5 \cos x-3 \sin x$,then $\frac{d^2 y}{d x^2}=$ . . . . . . .

If $y=e^{\tan ^{-1} x}$, then:

The second derivative of the function $y = f(x)$ is $f''(x) = 6(x - 1)$. If the graph of the function passes through the point $(2, 1)$ and the equation of the tangent at that point is $y = 3x - 5$,find the equation of the function.

Vedclass Test Series

Exam Paper Generator

Online Exam Module