If y is a function of x and log(x+y)=2xy,then | vedclass

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

Question

(C) Given equation: $\log(x+y)=2xy$ ... $(i)$ At $x=0$,$\log(0+y)=2(0)y$,which implies $\log(y)=0$,so $y=e^0=1$. Differentiating both sides of $(i)$ w

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Given equation: $\log(x+y)=2xy$ ... $(i)$ At $x=0$,$\log(0+y)=2(0)y$,which implies $\log(y)=0$,so $y=e^0=1$. Differentiating both sides of $(i)$ with respect to $x$: $\frac{1}{x+y} \left(1 + \frac{dy}{dx}\right) = 2y + 2x \frac{dy}{dx}$ Substitute $x=0$ and $y=1$ into the differentiated equation: $\frac{1}{0+1} \left(1 + \frac{dy}{dx}\right) = 2(1) + 2(0) \frac{dy}{dx}$ $1 + \frac{dy}{dx} = 2$ $\frac{dy}{dx} = 2 - 1 = 1$ Thus,the value of $\frac{dy}{dx}$ at $x=0$ is $1$.

If $y$ is a function of $x$ and $\log(x+y)=2xy$,then $\frac{dy}{dx}$ at $x=0$ is

Similar Questions

If $\sin(xy) + \cos(xy) = 0$,then $\frac{dy}{dx} = $

If $f(x) = \frac{1}{x^3} \int_5^x (2u^2 - u f'(u)) du$, then $f'(5) = $

If ${x^y} = {e^{x - y}}$,then $\frac{dy}{dx} = $

If $\sin ^2 x + \cos ^2 y = 1$,then $\frac{dy}{dx} = $

If $y$ is a function of $x$ and $\log (x+y)=2xy$,then the value of $y^{\prime}(0)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module