another implicit differentiation question

**differentiate** · January 5th 2010, 01:59 AM

Here's another:

If $ax^2 + by^2 = c$ show that the second derivative is equal to $-ac/b^2y^3$

Thanks

**Prove It** · January 5th 2010, 02:18 AM

Originally Posted by differentiate

Here's another:

If $ax^2 + by^2 = c$ show that the second derivative is equal to $-ac/b^2y^3$

Thanks

Impossible.

Taking the first derivative will eliminate the c. How can the second derivative have a c in it?

**HallsofIvy** · January 5th 2010, 02:50 AM

Originally Posted by Prove It

Impossible.

Taking the first derivative will eliminate the c. How can the second derivative have a c in it?

Because the second derivative will have an "x" in it and eliminating that reintroduces the "c".

Since $ax^2+ by^2= c$ , $2ax+ 2byy'= 0$ . Differentiating again, $2a+ 2by'^2+ 2byy"= 0$ .

$y"= \frac{-a- by'^2}{by}$ .

From $ax+ byy'= 0$ , $y'= -\frac{ax}{by}$ and so $y'^2= \frac{a^2x^2}{b^2y^2}$ .

Then $y"= -\frac{a+ \frac{a^2bx^2}{by^2}}{by}$
$y"= -\frac{aby^2+ a^2x^2}{b^2y^3}$

From $ax^2+ by^2= c$ , $ax^2= c- by^2$ so $a^2x^2= ac- aby^2$ .

$aby^2+ a^2x^2= aby^2+ ac- aby^2= ac$

$y"= -\frac{ac}{b^2y^3}$

**Shanks** · January 5th 2010, 02:56 AM

$2ax+2byy'=0,\text{ thus }y'=\frac{-ax}{by}<br /> 2a+2b(y')^2+2byy''=0$
combined with $y'=\frac{-ax}{by}$ ,gives
$y''=\frac{-ac}{b^2y^3}$ .

**Archie Meade** · January 5th 2010, 03:11 AM

Originally Posted by HallsofIvy

Because the second derivative will have an "x" in it

$Brilliant\ work\ !$

$(implied.....\ the\ 2nd\ derivative\ contains\ (y')^2\ which\ contains\ x^2)$

**differentiate** · January 5th 2010, 11:59 PM

Thanks

where does $(y')^2$ come from?

I get that you use the product rule, but how is that equal to $(y')^2$

**Archie Meade** · January 6th 2010, 03:43 PM

Originally Posted by HallsofIvy

Since $ax^2+ by^2= c$ , $2ax+ 2byy'= 0$ . Differentiating again, $2a+ 2by'^2+ 2byy"= 0$ .

1st derivative...

$\frac{d}{dx}(ax^2+by^2)=\frac{d}{dx}c$

$2ax+b\frac{d}{dx}y^2=0$

$2ax+b\frac{dy}{dx}\frac{d}{dy}y^2=0$

$2ax+2yb\frac{dy}{dx}=0$

2nd derivative...
2by is one factor and y' is the other of 2by(y'), or y and 2by' etc.

$2a+2by\frac{d}{dx}\frac{dy}{dx}+\frac{dy}{dx}2b\fr ac{dy}{dx}=0$

$2a+2by\frac{d^2y}{dx^2}+2b(\frac{dy}{dx})^2=0$

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