If {x^p}{y^q} = {(x + y)^{p + q}}, then frac{ | vedclass

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

Question

(A) Given the equation: ${x^p}{y^q} = {(x + y)^{p + q}}$Taking the natural logarithm on both sides:$\ln({x^p}{y^q}) = \ln({(x + y)^{p + q}})$$p\ln x +

Vedclass · Accepted Answer

$\frac{y}{x}$

Vedclass · Answer

(A) Given the equation: ${x^p}{y^q} = {(x + y)^{p + q}}$Taking the natural logarithm on both sides:$\ln({x^p}{y^q}) = \ln({(x + y)^{p + q}})$$p\ln x + q\ln y = (p + q)\ln(x + y)$Differentiating both sides with respect to $x$:$\frac{d}{dx}(p\ln x + q\ln y) = \frac{d}{dx}((p + q)\ln(x + y))$$\frac{p}{x} + \frac{q}{y}\frac{dy}{dx} = \frac{p + q}{x + y}(1 + \frac{dy}{dx})$Multiply by $xy(x + y)$ to clear denominators:$py(x + y) + qx(x + y)\frac{dy}{dx} = x(p + q)y + x(p + q)x\frac{dy}{dx}$$pxy + py^2 + qx^2\frac{dy}{dx} + qxy\frac{dy}{dx} = pxy + qxy + (px^2 + qx^2)\frac{dy}{dx}$Rearranging terms to isolate $\frac{dy}{dx}$:$\frac{dy}{dx}(qx^2 + qxy - px^2 - qx^2) = pxy + qxy - pxy - py^2$$\frac{dy}{dx}(qxy - px^2) = qxy - py^2$$\frac{dy}{dx} = \frac{qxy - py^2}{qx^2 - pxy} = \frac{y(qx - py)}{x(qx - py)}$$\frac{dy}{dx} = \frac{y}{x}$

If ${x^p}{y^q} = {(x + y)^{p + q}},$ then $\frac{dy}{dx} = $

Similar Questions

Evaluate $\int_{-2}^1 f(x) dx$,where $f(x)= \begin{cases} 1-2x, & x \leq 0 \\ 1+2x, & x > 0 \end{cases}$

Pulses are obtained from which family?

The equivalent capacitance between points $A$ and $B$ of the circuit shown will be

$10 \ g$ of a radioactive element is disintegrated to $1 \ g$ in $2.303 \ \text{minutes}$. What is the half-life (in minutes) of that radioactive element?

If the equilibrium constant for the reaction,$H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$ is $K$,what is the equilibrium constant of $HI_{(g)} \rightleftharpoons \frac{1}{2} H_{2(g)} + \frac{1}{2} I_{2(g)}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module