Let x^{k}+y^{k}=a^{k} where a, k 0. If frac | vedclass

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

Question

(C) Given the equation $x^{k}+y^{k}=a^{k}$.Differentiating both sides with respect to $x$,we get:$k x^{k-1} + k y^{k-1} \frac{dy}{dx} = 0$.Dividing by

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(C) Given the equation $x^{k}+y^{k}=a^{k}$.Differentiating both sides with respect to $x$,we get:$k x^{k-1} + k y^{k-1} \frac{dy}{dx} = 0$.Dividing by $k$ (since $k > 0$):$x^{k-1} + y^{k-1} \frac{dy}{dx} = 0$.Rearranging to solve for $\frac{dy}{dx}$:$\frac{dy}{dx} = -\frac{x^{k-1}}{y^{k-1}} = -\left(\frac{x}{y}\right)^{k-1}$.We are given $\frac{dy}{dx} + \left(\frac{y}{x}\right)^{\frac{1}{3}} = 0$,which implies $\frac{dy}{dx} = -\left(\frac{y}{x}\right)^{\frac{1}{3}} = -\left(\frac{x}{y}\right)^{-\frac{1}{3}}$.Comparing the two expressions for $\frac{dy}{dx}$:$-\left(\frac{x}{y}\right)^{k-1} = -\left(\frac{x}{y}\right)^{-\frac{1}{3}}$.Equating the exponents:$k - 1 = -\frac{1}{3}$.Solving for $k$:$k = 1 - \frac{1}{3} = \frac{2}{3}$.

Let $x^{k}+y^{k}=a^{k}$ where $a, k > 0$. If $\frac{dy}{dx}+\left(\frac{y}{x}\right)^{\frac{1}{3}}=0$,then the value of $k$ is:

Similar Questions

Find $\frac{dy}{dx}$ if $2x^2 - 3xy + y^2 + x + 2y - 8 = 0$.

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

If $3^x y^x = x^{3y}$, then the value of $\frac{dy}{dx}$ at $x = 1$ is:

On differentiation, if we obtain $f(x, y) dy - g(x, y) dx = 0$ from $2x^2 - 3xy + y^2 + x + 2y - 8 = 0$, then find the value of $\frac{g(2, 2)}{f(1, 1)}$.

If the tangent to the curve $xy + ax + by = 0$ at $(1, 1)$ makes an angle of $\tan^{-1} 2$ with the positive direction of the $x$-axis, then the value of $\frac{ab}{a + b}$ is...

Vedclass Test Series

Exam Paper Generator

Online Exam Module