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My problem is that I get two different answers when I use two different approaches to differentiate this problem with respect to $x$:

$$\left(\frac{3x+2y^2}{x^2+y^2}\right) =4$$

My first thought was to use the quotient rule, which gives me the answer

$$\frac{dy}{dx} = \frac{4x^2y+3x^2-3y^2}{4x^2y-6xy}.$$

However, someone showed me that if you first get rid of the rational expression like this:

$$3x+2y^2 = 4x^2+4y^2$$

and then differentiate, you get

$$\frac{dy}{dx}=\frac{8x-3}{4y}$$ (which is also the answer wolfram alpha gives).

I tried this also on a much simpler rational expression (to check that the problem wasn't just me beigng bad at algebra) and got different results again! Why?

Thanks for any help!

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  • $\begingroup$ It seems that you committed error in both calculations. $\endgroup$ – Nitin Uniyal Sep 23 '16 at 10:08
  • $\begingroup$ @mathlover What is the error in the first (quotient rule) calculation then? $\endgroup$ – StackTD Sep 23 '16 at 12:46
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Good news: the results may seem different, but aren't.

and then differentiate, you get $$\frac{dy}{dx}=\frac{8x-3}{4y}$$

There's a small (sign) mistake here, you should find: $$\frac{dy}{dx}=\frac{3-8x}{4y}$$

This may look different from the form you found via the quotient rule, but remember that you have a relation between $x$ and $y$, the original (implicit) function. So we hope the following equality holds:

$$\begin{array}{crcl} {} & \displaystyle \frac{4x^2y+3x^2-3y^2}{4x^2y-6xy} & = & \displaystyle \frac{3-8x}{4y} \\[5pt] {\Leftrightarrow} & 4y\left( 4x^2y+3x^2-3y^2 \right) & = & \left( 4x^2y-6xy \right) \left( 3-8x \right)\end{array}$$

Expanding, moving everything to the left-hand side and simplifying gives:

$$2 y \left( 16 x^3-24 x^2+4 x \cdot \color{blue}{2y^2}+9 x-3 \cdot \color{blue}{2y^2} \right)=0 \tag{$*$}$$

But from: $$\left(\frac{3x+2y^2}{x^2+y^2}\right) =4$$ we have that: $$3x+2y^2=4x^2+4y^2 \Rightarrow \color{blue}{2y^2=3x-4x^2}$$

Substitution into $(*)$ and simplifying, will give you $0$.

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