If {x^m}{y^n} = 2{(x + y)^{m + n}}, the value | vedclass

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

Question

(C) Given the equation: ${x^m}{y^n} = 2{(x + y)^{m + n}}$Taking the natural logarithm on both sides:$\ln({x^m}{y^n}) = \ln(2{(x + y)^{m + n}})$$m\ln x

Vedclass · Accepted Answer

$y/x$

Vedclass · Answer

(C) Given the equation: ${x^m}{y^n} = 2{(x + y)^{m + n}}$Taking the natural logarithm on both sides:$\ln({x^m}{y^n}) = \ln(2{(x + y)^{m + n}})$$m\ln x + n\ln y = \ln 2 + (m + n)\ln(x + y)$Now,differentiate both sides with respect to $x$:$\frac{d}{dx}(m\ln x + n\ln y) = \frac{d}{dx}(\ln 2 + (m + n)\ln(x + y))$$\frac{m}{x} + \frac{n}{y}\frac{dy}{dx} = 0 + \frac{m + n}{x + y}(1 + \frac{dy}{dx})$$\frac{m}{x} + \frac{n}{y}\frac{dy}{dx} = \frac{m + n}{x + y} + \frac{m + n}{x + y}\frac{dy}{dx}$Rearranging the terms to isolate $\frac{dy}{dx}$:$\frac{dy}{dx}(\frac{n}{y} - \frac{m + n}{x + y}) = \frac{m + n}{x + y} - \frac{m}{x}$$\frac{dy}{dx}(\frac{nx + ny - my - ny}{y(x + y)}) = \frac{mx + nx - mx - my}{x(x + y)}$$\frac{dy}{dx}(\frac{nx - my}{y(x + y)}) = \frac{nx - my}{x(x + y)}$$\frac{dy}{dx} = \frac{nx - my}{x(x + y)} 	imes \frac{y(x + y)}{nx - my}$$\frac{dy}{dx} = \frac{y}{x}$

If ${x^m}{y^n} = 2{(x + y)^{m + n}},$ the value of $\frac{dy}{dx}$ is

Similar Questions

If $y=x^{\sin x}+(\sin x)^x$,then $\frac{d y}{d x}$ at $x=\frac{\pi}{2}$ is

If $y = \left(\frac{x^{2}}{x+1}\right)^{x}$ and $\frac{dy}{dx} = y \left[g(x) + \log \left(\frac{x^{2}}{x+1}\right)\right]$,then $g(x) =$

If $h(x) = x^{x^x}$,then at $x = 1$,$\frac{h'(x)}{h(x)}$ is equal to

If $y = \sin x \cdot \sin 2x \cdot \sin 3x \cdot \ldots \cdot \sin nx$,then $y^{\prime}$ is

Differentiate the function with respect to $x$: $(\sin x - \cos x)^{(\sin x - \cos x)}$,where $\frac{\pi}{4} < x < \frac{3\pi}{4}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module