Question

2x^2+y^2=8xy, find the dy/dx.

Answer 1

Answer:

$\frac{dy}{dx}=\frac{4y-2x}{y-4x}$

Step-by-step explanation:

$\frac{d}{dx}(2x^2)=4x$

$\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$

$\frac{d}{dx}(8xy)$

$=8\frac{d}{dx}(xy)$

$=8(\frac{d}{dx}(x)y+x\frac{d}{dx}(y))$

$=8[1y+x\frac{dy}{dx}]$

$=8y+8x\frac{dy}{dx}$

Let's put it altogether now:

$2x^2+y^2=8xy$

Differentiating both sides gives:

$4x+2y\frac{dy}{dx}=8y+8x\frac{dy}{dx}$

We are solving for dy/dx so we need to gather those terms on one side and the terms without on the opposing side:

I'm going to first subtract 4x on both sides:

$2y\frac{dy}{dx}=8y-4x+8x\frac{dy}{dx}$

I'm not going to subtract 8xdy/dx on both sides:

$2y\frac{dy}{dx}-8x\frac{dy}{dx}=8y-4x$

It is time to factor the dy/dx out of the two terms on the left:

$\frac{dy}{dx}(2y-8x)=8y-4x$

Divide both sides by (2y-8x):

$\frac{dy}{dx}=\frac{8y-4x}{2y-8x}$

Reduce right hand side fraction:

$\frac{dy}{dx}=\frac{4y-2x}{y-4x}$

Answer 2

Answer:

$\frac{8y-4x}{2y-8x}$

Step-by-step explanation:

Differentiate implicitly with respect to x

noting that

$\frac{d}{dx}$ (y² ) = 2y$\frac{dy}{dx}$

Differentiate 8xy using the product rule

Given

2x² + y² = 8xy, then

4x + 2y$\frac{dy}{dx}$ = 8x$\frac{dy}{dx}$ + 8y

Collect terms in $\frac{dy}{dx}$

2y$\frac{dy}{dx}$ - 8x$\frac{dy}{dx}$ = 8y - 4x

$\frac{dy}{dx}$ (2y - 8x) = 8y - 4x

$\frac{dy}{dx}$ = $\frac{8y-4x}{2y-8x}$