Proof of Implicit Differentiation (showing a statement is true)

Question

Prove that if $x^2+y^2-2y\sqrt{1+x^2} = 0$, then $dy/dx = x/\sqrt{1+x^2}$.

Whilst I have implicitly differentiated in terms of x in order to derive that

$$dy/dx = (-x+2xy/\sqrt{1+x^2})/(y\sqrt{1+x^2}-1-x^2)$$

however I am unsure as to what my next steps are. I believe that I will need to rearrange the original supposition in order to achieve the proof however I do require some help as to how can I do this as I cannot see what would possibly cancel out.

Additionally, does anyone have any particularly handy methods/techniques to be able to complete proofs of this format of question more easily? I do understand that there is not one singular technique that can be used when completing this proofs but is there any strategy that minimizes the number of dead-ends that I come across in my working?

Thanks

Answer 1

A simpler way: the identity $$x^2+y^2-2y\sqrt{1+x^2} = 0$$ is equivalent to $$y^2+(1+x^2)-2y\sqrt{1+x^2} = 1$$ that is $$(y-\sqrt{1+x^2})^2=1$$ and differentiating in terms of $x$ both sides, we obtain $$2(y-\sqrt{1+x^2})\left(y'-\frac{x}{\sqrt{1+x^2}}\right)=0.$$ The first factor is always different from zero: in fact letting $y=\sqrt{1+x^2}$ into the given equation we get $$x^2+(1+x^2)-2(1+x^2) = 0$$ that is $-1=0$ which is impossible. Therefore the second factor has to be zero, and we find $$y'=\frac{x}{\sqrt{1+x^2}}.$$

Preliminary step. Notice that by the two dimensional implicit function theorem, $y'(x)$ exists (and therefore we are allowed to take the derivative in terms of $x$) if $$\frac{\partial F}{\partial y}=2(y-\sqrt{1+x^2})\not=0$$ where $F(x,y)=x^2+y^2-2y\sqrt{1+x^2}$, which holds as shown above.

Answer 2

Solve the quadratic of $y$ to get $$u=\sqrt{1+x^2}\pm 1 \implies \frac{dy}{dx}= \frac{x}{\sqrt{1-x^2}}.$$

Answer 3

I've a short trick for this.

For an implicit equation of $x$ and $y$ such that $f(x, y) = 0$, following result holds true: $$ \dfrac{dy}{dx} = \dfrac{-\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$$

For the given equation $x^2 + y^2 - 2y\sqrt{1+x^2} = 0$, derivative of $y$ with respect to $x$ is given by, $$\begin{aligned}\dfrac{dy}{dx} &= -\left[\dfrac{\frac{\partial }{\partial x}(x^2 + y^2 - 2y\sqrt{1+x^2} )}{\frac{\partial }{\partial y}(x^2 + y^2 - 2y\sqrt{1+x^2} )}\right]\\& = -\left[\dfrac{2x - 2y\cdot \frac{x}{\sqrt{1+x^2}}}{2y - 2\sqrt{1+x^2} }\right]\\& =-\left[\dfrac{x(\sqrt{1+x^2} - y)\cdot\frac1{\sqrt{1+x^2}}}{(y - \sqrt{1+x^2} )}\right] \\& =\frac{x}{\sqrt{1+x^2}}\end{aligned}$$

Answer 4

We find from the given relation $ \ x^2 + y^2 \ = \ 2y\sqrt{1+x^2} \ $ that

$$ 2x \ + \ 2y·\frac{dy}{dx} \ \ = \ \ 2·\frac{dy}{dx}·\sqrt{1+x^2} \ + \ 2y·\frac12·\frac{1}{\sqrt{1+x^2}}·2x \ $$ $$ \Rightarrow \ \ ( \ y \ - \ \sqrt{1+x^2} \ ) ·\frac{dy}{dx} \ \ = \ \ y· \frac{1}{\sqrt{1+x^2}}· x \ - \ x \ \ \Rightarrow \ \ \frac{dy}{dx} \ \ = \ \ \frac{-x \ + \ \frac{xy}{\sqrt{1+x^2}}}{y \ - \ \sqrt{1+x^2} } \ \ . $$ Replacing (at least for the present) the radical using the original equation $ ( \ \sqrt{1+x^2} \ = \ \frac{x^2 + y^2}{2y} \ ) \ $ , we obtain $$ \frac{dy}{dx} \ \ = \ \ \frac{ xy·\frac{2y}{x^2 + y^2} \ - \ x}{y \ - \ \frac{x^2 + y^2}{2y} } \ \ = \ \ \frac{ x·\left( \ 2y^2 \ - \ [x^2 + y^2] \ \right)}{y·(x^2 + y^2) \ - \ \frac{(x^2 + y^2)^2}{2y} } \ \ = \ \ \frac{ x· ( y^2 \ - \ x^2 )·2y}{2y^2·(x^2 + y^2) \ - \ (x^2 + y^2)^2 } $$ $$ = \ \ \frac{ x· ( y^2 \ - \ x^2 )·2y}{(2y^2 \ - \ x^2 \ - \ y^2)·(x^2 + y^2) } \ \ = \ \ \frac{ x· ( y^2 \ - \ x^2 )·2y}{( y^2 \ - \ x^2 )·(x^2 + y^2) } \ \ = \ \ x· \frac{ 2y}{ x^2 + y^2 } $$ $$= \ \ x· \frac{1}{ \sqrt{1+x^2} } \ \ , $$ again using the original relation. (So it appears there may be an error in your expression for the implicit derivative.)