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Given the formula $x^{2}y^{2}-8x=3$, find the second derivative.

I calculated the first derivative as $$-\frac{xy^{2}+4}{x^{2}y}$$

Working from that, I calculated the second derivative starting with $$\frac{([x^{2}y)\frac{d}{dx}(-xy^{2}+4)]-[(-xy^{2}+4)\frac{d}{dx}(x^{2}y)]}{(x^{2}y)^{2}}$$

The left $\frac{d}{dx}$ was calculated by $$\frac{d}{dx}(-xy^{2}+4)= [\frac{d}{dx}(-xy^{2})]+[\frac{d}{dx}(4)]= -x\frac{d}{dx}(y^{2})+ y^{2}\frac{d}{dx}(-x)= -2xy\frac{dy}{dx}- y^{2}$$

The right $\frac{d}{dx}$ was calculated by $$\frac{d}{dx}(x^{2}y)= x^{2}\frac{d}{dx}(y)+ y\frac{d}{dx}(x^{2})= x^{2}\frac{dy}{dx}+2xy$$

Plugging everything into the formula resulted in $$\frac{[(x^{2}y)(-2xy\frac{d}{dx}-y^{2}] - [(-xy^{2}+4)(x^2\frac{dy}{dx}+2xy)]}{(x^{2}y)^2}$$

$$\frac{[(-2x^{2}y^{2}\frac{dy}{dx}- 2x^{2}y^{3}]- [(x^{3}y^{2}\frac{dy}{dx}+2x^{2}y^{3}- 4x^2\frac{dy}{dx}-8xy]}{x^{4}y{2}}$$

Combining like terms resulted in


This is where I stall out. All my previous examples haven't been in quotient form, and how do I produce a $\frac{dy}{dx}$ from a quotient?

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Please check the first derivative. I found $\frac{dy}{dx}=-\frac{y^{2}x-4}{x^{2}y}$ – Américo Tavares Mar 8 '11 at 22:38
+1 for showing you are working hard at it before turning to the group, but do check your calculations on the derivative... – Arturo Magidin Mar 8 '11 at 23:43
Thanks, I try to work it out beforehand. Unfortunately, I am running up against some of the shortcomings of the school math dept: limited tutoring in fixed hours and a hard to follow professor. – Jason Mar 9 '11 at 1:27
up vote 3 down vote accepted

It's actually a bit simpler to work directly with the original (since then you don't have to worry about the quotient rule). Simply take derivatives twice, and then solve for $y''$ in terms of $x$, $y$, and $y'$; only then plug in $y'$.

Start with $$x^2y^2 - 8x=3.$$ Taking derivatives once, we get \begin{align*} \frac{d}{dx}\Bigl(x^2y^2 - 8x\Bigr) &= \frac{d}{dx}3\\ x^2\frac{d}{dx}y^2 + y^2\frac{d}{dx}x^2 - 8 &= 0\\ x^2\bigl(2yy'\bigr) + y^2\bigl(2x\bigr) -8 &= 0\\ 2x^2yy' + 2xy^2 - 8&= 0\\ x^2yy' + xy^2 - 4&= 0. \end{align*} Now take derivatives again, and solve for $y''$: \begin{align*} \frac{d}{dx}\Bigl(x^2yy' + xy^2-4\Bigr) &= 0\\ x^2y\left(\frac{d}{dx}y'\right) + x^2y'\left(\frac{d}{dx}y\right) + yy'\left(\frac{d}{dx}x^2\right)\\ \quad+x\left(\frac{d}{dx}y^2\right) + y^2\left(\frac{d}{dx}x\right) &=0\\ x^2yy'' + x^2(y')^2 + 2xyy' + 2xyy' + y^2 &=0\\ x^2yy'' +x^2(y')^2 + 4xyy' + y^2 &=0\\ x^2yy'' &= -\Bigl( x^2(y')^2 + 4xyy' + y^2\Bigr)\\ y''&= -\frac{x^2(y')^2 + 4xyy' + y^2}{x^2y}\\ y''&= -\frac{(y')^2}{y} - \frac{4y'}{x} - \frac{y}{x^2}. \end{align*}

If you want to get the value entirely in terms of $x$ and $y$, you can go back to the formula we had with the first derivative, $$x^2yy' + xy^2 - 4 = 0,$$ we can solve for $y'$ to get \begin{align*} x^2yy' &= 4 - xy^2\\ y' &= \frac{4-xy^2}{x^2y}\\ y' &= \frac{4}{x^2y} - \frac{y}{x}. \end{align*} So we can plug in this value of $y'$ into the formula for $y''$: \begin{align*} y'' &= -\frac{(y')^2}{y} - \frac{4y'}{x} - \frac{y}{x^2}\\ y'' &= -\frac{1}{y}\left(\frac{4}{x^2y} - \frac{y}{x}\right)^2 - \frac{4}{x}\left(\frac{4}{x^2y}-\frac{y}{x}\right) - \frac{y}{x^2}\\ y'' &= -\frac{1}{y}\left(\frac{16}{x^4y^2} - \frac{8y}{x^3y} + \frac{y^2}{x^2}\right)-\frac{16}{x^3y} + \frac{4y}{x^2} - \frac{y}{x^2}\\ y'' &= -\frac{16}{x^4y^3} + \frac{8}{x^3y} - \frac{y}{x^2} - \frac{16}{x^3y} + \frac{3y}{x^2}\\ y'' &= -\frac{16}{x^4y^3} - \frac{8}{x^3y}+ \frac{2y}{x^2}. \end{align*}

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So all your work has found that $$ \frac{d^2y}{dx^2}=\frac{-x^{3}y^{2}\frac{dy}{dx}-4x^{2}\frac{dy}{dx}-8xy}{x^{4}y^{3}} $$ but you have that $$ \frac{dy}{dx}=-\frac{xy^2+4}{x^2y} $$ so you can substitute that in to find the second derivative in terms of $x$ and $y$.

However, I think the first derivative is slightly different from what you have. $$ \frac{d}{dx}[x^2]y^2+x^2\frac{d}{dx}[y^2]-\frac{d}{dx}[8x]=\frac{d}{dx}[3] $$ gives $$ 2xy^2+x^2\cdot 2y\frac{dy}{dx}-8=0\implies \frac{dy}{dx}=\frac{xy^2-4}{-x^2y} $$

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but I have to add 8 to itself on both the left and right hand sides to move it over to the left side of equation, and thats where the +4 comes from after factoring. Also, what do you mean by substituting? Are you saying to replace $\frac{dy}{dx} with the first derivative?$ – Jason Mar 8 '11 at 22:36
@Jason, the issue is not the $4$, it's the fact that $$\frac{xy^2-4}{-x^2y}=-\frac{xy^2-4}{x^2y}=\frac{-xy^2+4}{x^2y}$$ so I think you forgot to pull a negative sign from the $4$. Americo Tavares' comment shows that he got the same first derivative that I did. And yes, you can replace $\frac{dy}{dx}$ with the first derivative to get the second derivative entirely in terms of $x$ and $y$. Of course, you'll want to recalculate with the correct first derivative. – yunone Mar 8 '11 at 22:46
I went through my calculations again, and matched up with what you wrote. My error was transforming the negative numerator to apply to the $\frac{dy}{dx}$ as a whole – Jason Mar 9 '11 at 1:31
@Jason, yeah, just a small algebra problem. The rest of your method works fine after that. – yunone Mar 9 '11 at 2:58

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