Implicit function the right approach

Question

Given $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ what is $y’_x$?

I’ve gone back and forth on this and I thought I could perhaps use the implicit function theorem, but then again is there a need to? I have the answer options but can’t seem to get to them so I must be doing something wrong. Not sure if it’s from the algebra or my approach.

Answer 1

You have ${x^2 \over a^2}+ {y^2 \over b^2} = 1$, hence $y = \pm b \sqrt{1-{x^2 \over a^2}}$.

We can define two functions $y_-(x) = - b \sqrt{1-{x^2 \over a^2}}$, $y_+(x) = b \sqrt{1-{x^2 \over a^2}}$ defined on $[-a,a]$ and differentiable on $(a,-a)$, with

$y_-'(x) = {xb \over a^2} {1 \over \sqrt{1-{x^2 \over a^2}}} = -{b^2 \over a^2} {x \over y_-(x)}$, $y_+'(x) = -{xb \over a^2} {1 \over \sqrt{1-{x^2 \over a^2}}} = = -{b^2 \over a^2} {x \over y_+(x)}$.

Answer 2

Hint:

Surely $(x^2)’ =2x$, $(1)’=0$, and for $(y^2)’$ use the chain rule.

It may help to think of $y$ as a function $y(x)$, so that term becomes $([y(x)]^2)’$.

Answer 3

I've said this before in other answers, but we can encapsulate our relation as a two variable function $$F(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ It is true for any two variable function that $$\frac{\mathrm{d} f}{\mathrm{d} u}=\frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}u}+\frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}u}$$ In the special case that $u=x$, $$\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}x}$$ Applying this to our example, $$\frac{\mathrm{d}}{\mathrm{d}x}F(x,y)=\frac{\mathrm{d}}{\mathrm{d}x}(1)=0= \frac{2x}{a^2}+\frac{2y}{b^2}\frac{\mathrm{d}y}{\mathrm{d}x}$$ Therefore $$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{-x}{y}\left(\frac{b}{a}\right)^2.$$