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 $y = ({x^x})^x$,then $\frac{dy}{dx} =$

If $x^{m} y^{n}=(x+y)^{m+n}$,then $\frac{dy}{dx}$ is equal to

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

If $y(x) = x^x, x > 0$,then $y^{\prime \prime}(2) - 2y^{\prime}(2)$ is equal to:

Differentiate the function with respect to $x$: $(x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module