# Implicit Differentiation

• Dec 27th 2007, 03:39 AM
slevvio
Implicit Differentiation
$\displaystyle x^2 + 4y^2 = 9$

Show that $\displaystyle \frac{d^2y}{dx^2} = \frac{45}{2(x+y)^3}$.

I was wondering if I could get some help with it. I get the derivate to be $\displaystyle -\frac{x}{4y}$ (which is correct) but when I try to differentiate this again I get a completely different answer to that in the question. Thanks, any help would be much appreciated.
• Dec 27th 2007, 05:51 AM
TKHunny
Does your answer still have dy/dx in it? Substituting the known expression for that, followed by a little algebra, might fix it up for you.

Does your expression still have x^2 + 4y^2 in it? Substituting the known value for that might fix it up for you.

Did you find the second derivative by differentiating the expression for the first derivative, using the quotient rule? Sometimes, it is more convenient to differentiate the original expression twice, using the product rule.

Just some hints. Without you showing your work, it's hard to know where you may have erred, if, indeed, you did.
• Dec 27th 2007, 05:57 AM
Isomorphism
Quote:

Originally Posted by TKHunny
Does your answer still have dy/dx in it? Substituting the known expression for that, followed by a little algebra, might fix it up for you.

Does your expression still have x^2 + 4y^2 in it? Substituting the known value for that might fix it up for you.

I tried it for a long time, and did not get the algebraic manipulation. Do you have any lead on the problem:confused:

I am beginning to think the problem is incorrect :(
• Dec 27th 2007, 06:37 AM
Soroban
Hello, slevvio!

I don't get their answer either . . .

Quote:

$\displaystyle {\color{blue} x^2 + 4y^2 \:= \:9}$

$\displaystyle \text{Show that: }\:\frac{d^2y}{dx^2} \:= \:\frac{45}{2(x+y)^3}$

Are you sure you copied this correctly?
This is an answer to an entirely different problem.

$\displaystyle \text{Differentiate implicitly: }\;2x + 8y\frac{dy}{dx}\:=\:0\quad\Rightarrow\quad\frac{dy }{dx}\:=\:-\frac{x}{4y}\:=\:-\frac{1}{4}\cdot\frac{x}{y}$

$\displaystyle \text{Differentiate: }\;\frac{d^2y}{dx^2} \;=\;-\frac{1}{4}\left[\frac{y\!\cdot\!1 - x\!\cdot\!\frac{dy}{dx}}{u^2}\right] \;=\;-\frac{1}{4}\cdot\frac{y - x\frac{dy}{dx}}{y^2}$

$\displaystyle \text{Since }\frac{dy}{dx} \:=\:-\frac{x}{4y}\text{, substitute: }\;\frac{d^2y}{dx^2} \;=\;-\frac{1}{4}\cdot\frac{y - x\left(-\frac{x}{4y}\right)}{y^2}$

$\displaystyle \text{Multiply top and bottom by }4y\!:\;\;\frac{d^2y}{dx^2}\;=\;-\frac{1}{4}\cdot\frac{4y^2 + x^2}{4y^3}$

$\displaystyle \text{So we have: }\;\frac{d^2y}{dx^2} \;=\;- \frac{{\color{blue}x^2+4y^2}}{16y^3}$

$\displaystyle \text{According to the original equation: }\:x^2+4y^2 \:=\:{\color{blue}9}$

$\displaystyle \text{Therefore: }\:\frac{d^2y}{dx^2} \:=\:-\frac{9}{16y^3}$

• Dec 27th 2007, 06:53 AM
Isomorphism
Quote:

Originally Posted by Soroban
Hello, slevvio!

I don't get their answer either . . .

$\displaystyle \text{Differentiate implicitly: }\;2x + 8y\frac{dy}{dx}\:=\:0\quad\Rightarrow\quad\frac{dy }{dx}\:=\:-\frac{x}{4y}\:=\:-\frac{1}{4}\cdot\frac{x}{y}$

$\displaystyle \text{Differentiate: }\;\frac{d^2y}{dx^2} \;=\;-\frac{1}{4}\left[\frac{y\!\cdot\!1 - x\!\cdot\!\frac{dy}{dx}}{u^2}\right] \;=\;-\frac{1}{4}\cdot\frac{y - x\frac{dy}{dx}}{y^2}$

$\displaystyle \text{Since }\frac{dy}{dx} \:=\:-\frac{x}{4y}\text{, substitute: }\;\frac{d^2y}{dx^2} \;=\;-\frac{1}{4}\cdot\frac{y - x\left(-\frac{x}{4y}\right)}{y^2}$

$\displaystyle \text{Multiply top and bottom by }4y\!:\;\;\frac{d^2y}{dx^2}\;=\;-\frac{1}{4}\cdot\frac{4y^2 + x^2}{4y^3}$

$\displaystyle \text{So we have: }\;\frac{d^2y}{dx^2} \;=\;- \frac{{\color{blue}x^2+4y^2}}{16y^3}$

$\displaystyle \text{According to the original equation: }\:x^2+4y^2 \:=\:{\color{blue}9}$

$\displaystyle \text{Therefore: }\:\frac{d^2y}{dx^2} \:=\:-\frac{9}{16y^3}$

• Dec 27th 2007, 08:21 AM
TKHunny
Well, I was hoping the student would show a little fortitude and tell me that the book is wrong. Oh, well.

The sign is wrong. The structure is wrong. I have to wonder if two problems aren't mixed up.
• Dec 27th 2007, 08:37 AM
slevvio
Thanks everybody my example sheet was clearly wrong the answers you stated are what I got as well.. thanks again
• Dec 27th 2007, 08:42 AM
slevvio
Quote:

Originally Posted by TKHunny
Well, I was hoping the student would show a little fortitude and tell me that the book is wrong. Oh, well.

I didn't know my book was wrong, hence the post.
• Dec 31st 2007, 09:36 AM
TKHunny
Ah! An opportunity to do a little soap-boxing.

1) Of course you don't know the book is wrong! This is the very strength of my point. If you have confidence in your results, you will discover the book wrong and prove it so. If you knew it was wrong, why would that require fortitude?

2) It is unfortunate that honesty occasionally is perceived as rudeness. Again, a little personal strength will fix this misconception. I'm not here to hurt anyone's feelings, but I do tend to tell the truth.

Personal Example: My first math exam in 7th grade, quite a few years ago. I managed a disappointing 94%, but I was pleased compared to the teacher's mere 96%. My exam was originally returned with a 90% marked on it. After some thought and further examination, I showed the teacher his two errors. It's not magic. It's just a little confidence.

3) In this case, I was only just barely talking about the student. I have a long-standing disagreement with Soroban's style of providing complete, worked examples rather than asking leading questions and encouraging the student to think things through. It's not a personal problem, it's just a significant style difference. This isn't news to Soroban. Normally, I just let it go, but once in a while I'll throw in a comment. To Soroban's credit, no one EVER accuses him of being rude. One must make choices in life. We need all kinds of teachers. I don't mind that some don't agree with my more aggressive style.