Given the function $y=\frac{m}{n}\sqrt{n^2-x^2}$, prove that $y′=\frac{-m^2x}{n^2y}$

Question

I am trying to solve an exercise with derivatives. As the tittle says, I have the following function:

$$y=\frac{m}{n}\sqrt{n^2-x^2}$$

$m$ and $n$ are constants. After finding the first derivative, I need to prove that:

$$y′=\frac{-m^2x}{n^2y}$$

You must have noticed that there is a $y$ in the answer, but you do not need to find the implicit derivative to get the answer. It is supposed that finding the first function somewhere in the first derivative and replacing it with $y$ you get that answer.

I already solved another exercise of this kind, so you can better understand how I am supposed to solve this one.

Example: Given the function $y=\sqrt{2mx}$, prove that $y′=\frac{m}{y}$

$$y=\sqrt{2mx}$$ $$y′=\frac{1}{2\sqrt{2mx}}\left(2m\right)$$ $$y′=\frac{m}{\sqrt{2mx}}$$ Since $y=\sqrt{2mx}$: $$y′=\frac{m}{y}$$

Answer 1

$y = \dfrac{m}{n}\sqrt{n^2 - x^2} = \dfrac{m}{n} (n^2 - x^2)^{1/2}; \tag 1$

$y' = \dfrac{m}{n} \left ( (-2x) \dfrac{1}{2}(n^2 - x^2)^{-1/2} \right ) = \dfrac{m}{n} (-x(n^2 - x^2)^{-1/2}) = -\dfrac{mx}{n} (n^2 - x^2)^{-1/2}; \tag 2$

we notice that (1) implies

$\dfrac{1}{y} = \dfrac{n}{m}(n^2 - x^2)^{-1/2}, \tag 3$

whence

$(n^2 - x^2)^{-1/2} = \dfrac{m}{ny}; \tag 4$

we insert (4) into (2):

$y' = -\dfrac{mx}{n} \dfrac{m}{ny} = -\dfrac{m^2x}{n^2y}, \tag 5$

as per request.

Answer 2

Well, the derivative of $y$ is $$y'=\frac{m}{n}\times(-2x)\times\frac{1}{2\sqrt{n^2-x^2}}=-\frac{mx}{n\sqrt{n^2-x^2}}$$

Then, note that $\frac{1}{\sqrt{n^2-x^2}}=\frac{m}{ny}$ so substituing, we get the desired result $$y'=-\frac{m^2x}{n^2y}$$

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