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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

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

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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    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 fi$x$ed hours and a hard to follow professor.2011-03-09

2 Answers 2

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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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    @Jason, yeah, just a small algebra problem. The rest of your method works fine after that.2011-03-09