Mathematical Analysis I - 2-3-1.(5)

Calculus Level 1

Given that x + x y + y = 0 x+\sqrt{xy}+y=0 ( x , y ∈ R ) (x,y \in \mathbb R) , and x x is independent variable, find d y d x \dfrac{dy}{dx} at x = 0 x=0 (if exists).

Note: This is an unintended great problem of our textbook.

− 1 -1 0 0 1 1 It does not exist. \textup{It does not exist.}

Alice Smith by Alice Smith , 20, China

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1 solution

Chris Lewis
Nov 11, 2020

Rearranging, x + y = − x y x+y=-\sqrt{xy} Squaring both sides and cancelling, we get x 2 + x y + y 2 = 0 x^2+xy+y^2=0

But x 2 + x y + y 2 = ( x + y 2 ) 2 + 3 y 2 4 x^2+xy+y^2=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}

As squares of real numbers, both of these terms are greater than or equal to zero; so we find the only real solution is x = y = 0 x=y=0 , and the derivative is not defined.

5 Helpful 2 Interesting 3 Brilliant 1 Confused

Hmph. Is there any way to define d y d x \dfrac{dy}{dx} in this case?

By the way, in light of my original question, do you have any ideas on an algebraic proof on the infinite sum I mentioned? (Hopefully you will remember me.)

Inquisitor Math - 7 months ago

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I can't think of a definition that would work; derivatives always involve limits, or rates of change, so if there's just a single point a derivative wouldn't make sense. The "tangent" to the graph could have any gradient.

I've been doing some of your other problems, but not looked at the sum one recently. I'll check back at some point (though probably not today).

Chris Lewis - 7 months ago

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Kay. Good to know!

Inquisitor Math - 7 months ago

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