0

I can find the derivative of the above using implicit differentiation but was interested in finding it using the explicit function way.

Nudged by “dx”:

$y=(25-x^2)^{1/2}\\$

$dy= (25-(x+dx)^2)^{1/2} - (25-x^2)^{1/2}$

I am unable to solve it beyond a few points. I am curious as to how this can be solved this way. I know to do it using implicit differentiation but like to see it solved this way. so far no luck. Any help is appreciated.

thanks.

Edit 1:

I was not clear.

From the above We get the following;

$dy= (25-x^2-2xdx)^{1/2} - (25-x^2)^{1/2}$

How to proceed beyond this to get the answer?

Thanks.

Khanna111
  • 111

6 Answers6

1

If I didn't do anything silly in my derivation, \begin{align} x^2 + y^2 &= 25\\ \therefore y &= \pm \sqrt{25-x^2}\\ \therefore \frac{dy}{dx} &= \frac{d}{dx}\left(\pm \sqrt{25-x^2}\right)\\ &= \pm \frac{-2x}{2\sqrt{25-x^2}}\\ &= \pm \frac{x}{\sqrt{25-x^2}} \end{align}

There are actually two solutions, of course, because $y$ is not a function of $x$ (it does not pass the 'vertical line test') so we may consider the derivative of the top half of the circle and the derivative of the bottom half of the circle separately.

Jack Crawford
  • 999
  • 1
    Better write $\mp$ than $\pm$ in the last line. –  Oct 28 '19 at 08:05
  • I saw this in your answer and was surprised: I have never seen $\mp$ before. In this context I'm not completely sure that it matters since of course there is no canonical ordering to the solutions, but thankyou for introducing me to this symbol. You're probably right that it adds a little bit to the clarity. – Jack Crawford Oct 28 '19 at 08:08
  • The thing is that $\pm(-a)=\pm a$ is not considered valid, the "$\pm$" remain "synchronized". Anyway, expressions like $\pm a\pm b$ are somewhat ambiguous. You see examples here: https://math.stackexchange.com/questions/866331/numerically-stable-algorithm-for-solving-the-quadratic-equation-when-a-is-very –  Oct 28 '19 at 08:22
  • Thanks. Just made question clearer. please take a look – Khanna111 Oct 28 '19 at 08:39
1

The is no special difficulty in $$y'(x)=\left(\pm\sqrt{25-x^2}\right)'=\mp\frac x{\sqrt{25-x^2}}.$$

If you want to reinvent the wheel,

$$\lim_{h\to0}\frac{\sqrt{25-(x+h)^2}-\sqrt{25-x^2}}h=\lim_{h\to0}\frac{x^2-(x+h)^2}{h(\sqrt{25-(x+h)^2}+\sqrt{25-x^2})} \\=\frac1{2\sqrt{25-x^2}}\lim_{h\to0}\frac{-2xh-h^2}h=-\frac x{\sqrt{25-x^2}}.$$

This can be written

$$y'(x)=-\frac{x}{y(x)},$$ which matches

$$2x+2y(x)y'(x)=(x^2+y^2(x))'=0.$$

0

We have that

$$y=(25-x^2)^{1/2} \implies dy =\frac12(25-x^2)^{-1/2}(-2x) dx=-\frac{x}{(25-x^2)^{1/2}}dx$$

user
  • 162,563
0

$x^2+y^2=25$,consider a very magnified neighbourhood of the curve at any point excepts the points $x=5,-5$.In that neighbourhood there can be defined as function $f$ such that $y=f(x)$ where $y$ satisfies the given relation.Now express $y$ in terms of $x$.Take one of $f(x)=-\sqrt{(25-x^2)}$or $f(x)=\sqrt{25-x^2}$.then differentiate like a function,you will ultimately get the same result.

Kishalay Sarkar
  • 4,238
0

For your method to work, you would need the best linear approximation of $x \mapsto \sqrt{a+x}$ at $x=0$, which is an equivalent definition of the derivative at $x=0$.

Botond
  • 12,134
0

Use the chain rule since $f=f(x,y)$:

$f(x,y)=x^2+y^2-25=0$

$\frac{d}{dx}{f(x,y)}=2x+\frac{d}{dx}{(y^2)}=2x+\frac{d}{dy}{(y^2)}\frac{dy}{dx}=2x+2y\frac{dy}{dx}=0$

$2x+2y\frac{dy}{dx}=2x+ 2\sqrt{25-x^2} \frac{dy}{dx}=0$

Lele Mabur
  • 23
  • 5