Question

If x^2+xy+y^2=0 then find dy/dx

Answer 1

Answer:

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Answer 2

$\boxed{\dfrac{dy}{dx}=-\dfrac{2x+y}{x+2y}}$

Given data:

The equation $x^{2}+xy+y^{2}=0$

To find:

$\dfrac{dy}{dx}$

Concept to be used:

• If $u,v$ are the functions of $x$, then

$\dfrac{d}{dx}{uv}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$

$\dfrac{d}{dx}(u\pm v)=\dfrac{du}{dx}\pm\dfrac{dv}{dx}$

• $\dfrac{d}{dx}(x^{n})=n\:x^{n-1}$, where $n$ is a rational number.

Step-by-step explanation:

Given, $x^{2}+xy+y^{2}=0$

Differentiating both sides with respect to $x$, we get

$\Rightarrow \dfrac{d}{dx}(x^{2}+xy+y^{2})=\dfrac{d}{dx}(0)$

$\dfrac{d}{dx}(x^{2})+\dfrac{d}{dx}(xy)+\dfrac{d}{dx}(y^{2})=0$

$\Rightarrow 2x+x\dfrac{dy}{dx}+y+2y\dfrac{dy}{dx}=0$

$\Rightarrow (x+2y)\dfrac{dy}{dx}=-2x-y$

$\Rightarrow \dfrac{dy}{dx}=-\dfrac{2x+y}{x+2y}$

#SPJ3