Here's another:

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

Thanks

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- Jan 5th 2010, 01:59 AMdifferentiateanother implicit differentiation question
Here's another:

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

Thanks - Jan 5th 2010, 02:18 AMProve It
- Jan 5th 2010, 02:50 AMHallsofIvy
Because the second derivative will have an "x" in it and eliminating that reintroduces the "c".

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

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

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

Then $\displaystyle y"= -\frac{a+ \frac{a^2bx^2}{by^2}}{by}$

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

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

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

$\displaystyle y"= -\frac{ac}{b^2y^3}$ - Jan 5th 2010, 02:56 AMShanks
$\displaystyle 2ax+2byy'=0,\text{ thus }y'=\frac{-ax}{by}

2a+2b(y')^2+2byy''=0$

combined with $\displaystyle y'=\frac{-ax}{by}$,gives

$\displaystyle y''=\frac{-ac}{b^2y^3}$. - Jan 5th 2010, 03:11 AMArchie Meade
- Jan 5th 2010, 11:59 PMdifferentiate
Thanks

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

I get that you use the product rule, but how is that equal to $\displaystyle (y')^2 $ - Jan 6th 2010, 03:43 PMArchie Meade
1st derivative...

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

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

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

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

2nd derivative...

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

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

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