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!

  • $\begingroup$ 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

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 $0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.