Ian Lindsey

2023-03-23

What is 2xy differentiated implicitly?

Santino Potter

Beginner2023-03-24Added 8 answers

I'll assume that since the question doesn't say with respect to what, y is a function of x.

Use the product rule:

$y\prime =d\frac{(u.v)}{dx}=v.d\frac{u}{dx}+u.d\frac{v}{dx}$

So:

$y\prime =2x.y\prime +y2.\frac{dx}{dx}$

$y\prime =2x.y\prime +2y$

$y\prime =\frac{2y}{1-2x}$

Use the product rule:

$y\prime =d\frac{(u.v)}{dx}=v.d\frac{u}{dx}+u.d\frac{v}{dx}$

So:

$y\prime =2x.y\prime +y2.\frac{dx}{dx}$

$y\prime =2x.y\prime +2y$

$y\prime =\frac{2y}{1-2x}$

punctator2ueu

Beginner2023-03-25Added 8 answers

The function is

$f(x,y)=2xy$

Partially derivatives include

$\frac{\partial f}{\partial x}=2y$

$\frac{\partial f}{\partial y}=2x$

Hence,

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

$f(x,y)=2xy$

Partially derivatives include

$\frac{\partial f}{\partial x}=2y$

$\frac{\partial f}{\partial y}=2x$

Hence,

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

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