Reciprocal derivative, inverse function

Question

$$ 2x^2 + 2y + 1 = 0 $$

Determine the values of $x,y$ for which $\frac{dy}{dx}$ and $\frac{dx}{dy}$ are reciprocals.

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I'm not too sure if I'm interpreting this correctly.

If I have some $(a, b)$ then $(b, a)$ is on the inverse function.

The inverse exists if the domain is interval $I_1 = (-\infty, 0]$ or $I_2 = [0, + \infty)$

Is it asking me for some $(a,b)$ on $f$ such that $f'(a)$

$$ 2x^2 + 2y + 1 = 0 $$

Solving for $y$

$$ y = -\frac{1}{2}(1 + 2x^2) $$

Then

$$ f' = -2x $$

So I want to find some point $(a, b)$ for which

$$ (f^{-1})'(b) = -\frac{1}{2a} $$

I'm not sure how to go about this

Answer 1

if you take the derivate of both sides with respect to $x$ (while considering $y$ as a function of $x$) a 1st time then with respect to $y$ (while considering $x$ as a function of $y$)

you get these two equations

$$\begin{align} \frac{dy}{dx} = -2x \\ \frac{dx}{dy} = -\frac{1}{2x} \end{align}$$

now if you see both of them as a function of $x$

you have that $\frac{dy}{dx}o\frac{dx}{dy}(x) =\frac{dx}{dy}o\frac{dy}{dx}(x) =1$ for every $x\neq0$

therefore they are almost always reciprocal (except when $x$ is zero)

Answer 2

from your equation we get $$y=\frac{1}{2}(-1-2x^2)$$ then we get $$\frac{dy}{dx}=-2x$$ and $$x=\pm\sqrt{\frac{1}{2}(-1-2y)}$$ then $$\frac{dx}{dy}=\pm\frac{1}{2}\left(\frac{1}{2}(-1-2y)\right)^{-1/2}\left(\frac{1}{2}\right)(-2)$$ then we have $$-2x=\pm2\sqrt{\frac{1}{2}(-1-2y)}$$ simplifying we get $$2x^2=-1-2y$$ and thus this is always true if $$-1-2y\geq 0$$

Answer 3

We can differentiate the equation as $$ d(2x^2 + 2y + 1) = d(0) $$

where $d$ is the differential operator. By using the properties of the differential and noting that $d(constant)=0$ and $df=f^{'} dx$ (assuming that $f$ is a function of $x$) then,

$$ d(2x^2 + 2y + 1) = d(0)=0 \rightarrow d(2x^2)+d(2y)+d(1)=0 \\ \rightarrow 4xdx+2dy+0=0 \rightarrow \frac{dy}{dx}=-2x, \frac{dx}{dy}=-1/2x $$

Since $\frac{dx}{dy}=-1/2x$ is not defined for $x=0$, then $\frac{dy}{dx}$ and $\frac{dx}{dy}$ arereciprocals except $x=0$. The corresponding $y$ can be found using the original equation: $2*0+2y+1=0\rightarrow y=-1/2$. Thus the derivatives are reciprocals of each other except $(0,-1/2)$