Derivative of (2/x) + (1/y) = 4 and

[autodidude]

Can you somehow show that the derivative of $$\frac{2}{x}+\frac{1}{y}=4$$ is the same as $$2y+x=4xy$$ if x≠0 and y≠0? Or is that not true?

 

[chiro]

Can you somehow show that the derivative of $$\frac{2}{x}+\frac{1}{y}=4$$ is the same as $$2y+x=4xy$$ if x≠0 and y≠0? Or is that not true?

Yes you can do this (provided your condition that you have non-zero values for y and x) since algebraically both are equivalent (again assuming non-zero values).

The rest is going to be implicit differentiation, but again you have to consider that there will be holes at x = 0 and also at y = 0 where the function is defined for either of these values.

 

[autodidude]

Thanks chiro. To do so, would you just differentiate each equation and manipulate one of the derivatives until it's equivalent to the other? (again, with x and y no equal to 0

 

[Dickfore]

You don't even have to do that much. just multiply the original equation by $x y$. You will get the second equation. Multiplying by $x y$ is justified only when:
$$x y \neq 0 \Leftrightarrow x \neq 0 \wedge y \neq 0$$

 

[Mark44]

Is this your question?

Can you somehow show that the derivative of $$\frac{2}{x}+\frac{1}{y}=4$$ is the same as the derivative of $$2y+x=4xy$$ if x≠0 and y≠0? Or is that not true?

As Dickfore already said, the two equations are equivalent, as long as x≠0 and y≠0, so implicit differentiation would produce the same values for dy/dx in both equations. Keep in mind that it makes no sense to talk about the "derivative of an equation." You can implicitly differentiate both sides of an equation with respect to some variable, though, and then solve algebraically for dy/dx (or whatever derivative you're looking for).

 

[autodidude]

Thanks guys