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## Confounding Calculus

Another fun Calculus problem we ran into the other day involved using the quotient rule versus using the product rule.  This problem also came from the Briggs and Cochran book in the section on Implicit Differentiation (pg. 162 #15).  So, the problem this time started out as

Use implicit differentiation to find $\frac {dy}{dx}$.

$x^3=\frac {x+y}{x-y}$

So, if we start out by differentiating the expressions on each side, we get

$3x^2=\frac {(x-y)(1+y')-(x+y)(1-y')}{(x-y)^2}$

If we expand and simplify a little –

$3x^2=\frac {(x+xy'-y-yy')-(x-xy'+y-yy')}{(x-y)^2}$

$3x^2=\frac {x+xy'-y-yy'-x+xy'-y+yy'}{(x-y)^2}$

$3x^2=\frac {2xy'-2y}{(x-y)^2}$

So then

$3x^2(x-y)^2=2xy'-2y$

and

$\frac{3x^2(x-y)^2+2y}{2x}=y'$

This is the answer that appears in the text – but, if we assume that $x\neq y$, then we can clear the denominator in the original problem and use implicit differentiation on the result.

$x^3=\frac {x+y}{x-y}$

$x^3(x-y)=x+y$

$x^4-x^3y=x+y$

Then, differentiating

$4x^3-(3x^2y+x^3y')=1+y'$

or

$4x^3-3x^2y-x^3y'=1+y'$

Then, to isolate $y'$:

$4x^3-3x^2y-1=x^3y'+y'$

$4x^3-3x^2y-1=y'(x^3+1)$

and

$\frac{4x^3-3x^2y-1}{x^3+1}=y'$

I spent some time playing around with these, and believe me, they are not the same.  I graphed them and here is what the first answer $\frac{3x^2(x-y)^2+2y}{2x}=y'$ looks like:

And here is the second version, $\frac{4x^3-3x^2y-1}{x^3+1}=y'$

I tried plugging values for $x$ and $y$ into these different expressions for the derivative of the same function to see if I got the same answer.  This led me to realize the solution to the conundrum!

I started by plugging in some simple values for $x$ and $y$ like $x=1, y=1$ but quickly realized that this wasn’t valid because $x\neq y$.  So then I tried $x=1, y=0$ and for these values, the two expressions were the same!

$\frac{3(1)(1-0)^2+2(0)}{2(1)}=y'$

$\frac{3}{2}=y'$

and

$\frac{4(1)^3-3(1)^2(0)-1}{(1)^3+1}=y'$

$\frac{4-1}{1+1}=y'$

$\frac{3}{2}=y'$

These are the same – because the point (1,0) satisfies the original function: $x^3=\frac {x+y}{x-y}$.

So I got out the Excel spreadsheet and found a series of points that satisfied the original function and tried a few of these points in the two expressions for the derivative and for each one the value of the two expressions is the same!

So next I was determined to show that assuming the original function was true would show the equality of the two different expressions for the derivative.

So, assuming $x^3=\frac {x+y}{x-y}$ we need to turn $\frac{4x^3-3x^2y-1}{x^3+1}=y'$ into $\frac{3x^2(x-y)^2+2y}{2x}=y'$.

First I substituted $\frac{x+y}{x-y}$ in for $x^3$, so:

$\frac{4x^3-3x^2y-1}{x^3+1}=y'$

$\frac{4(\frac{x+y}{x-y})-3x^2y-1}{\frac{x+y}{x-y}+1}=y'$

$\frac{4(\frac{x+y}{x-y})-3x^2y-1}{\frac{x+y}{x-y}+\frac{x-y}{x-y}}=y'$

or

$\frac{4(\frac{x+y}{x-y})-3x^2y-1}{\frac{2x}{x-y}}=y'$

Then I cleared the fraction in the numerator and denominator:

$(\frac{x-y}{x-y})\frac{4(\frac{x+y}{x-y})-3x^2y-1}{\frac{2x}{x-y}}=y'$

$\frac{4(x+y)-3x^2y(x-y)-1(x-y)}{2x}=y'$

OK, so this is

$\frac{3x^2y^2-3x^3y+4x+4y-x+y}{2x}=y'$

or

$\frac{3x^2y^2-3x^3y+3x+5y}{2x}=y'$

I saw that where I was headed with this had a $2y$ and a $3x^2$ in it, so I moved a few things around and ended up with:

$\frac{3x^2y^2-3x^3y+3(x+y)+2y}{2x}=y'$

Going back to the original function $x^3=\frac {x+y}{x-y}$ I saw that I could substitute $x^3(x-y)$ in for the $(x+y)$:

$\frac{3x^2y^2-3x^3y+3(x^3(x-y))+2y}{2x}=y'$

$\frac{3x^2y^2-3x^3y+3(x^4-x^3y)+2y}{2x}=y'$

$\frac{3x^2y^2-3x^3y+3x^4-3x^3y+2y}{2x}=y'$

Factor out $3x^2$ from the everything except the $2y$ and this is starting to look good:

$\frac{3x^2(y^2-xy+x^2-xy)+2y}{2x}=y'$

And there it is:

$\frac{3x^2(x-y)^2+2y}{2x}=y'$

– sigh of relief –