Implicit Differentiation (am I wrong or is the author wrong?)

[rayroshi]

Okay, I am trying to work out a problem (problem # 16.5) found at the bottom of page 249 of Michael Kelley's book, The Humongous Book of Calculus Problems, and I just can't see how he got the answer that he did. After going over it for waaay longer than I care to admit, I think that I found an error in his solution, but then who am I, an old guy trying to learn calculus on his own, to tell? Here's the problem: "Given y = 9x^1/2 - 2y^3/5, find dy/dx." On the next page, p.250, he shows how he solves the problem, but it seems to me that he is wrong in his very first step, where he says, "Differentiate each term with respect to x," and then proceeds as follows: dy/dx = 9*1/2 (x^-1/2) - 2*3/5 (y^-2/5) * dy/dx (no problem, so far)...which all equals (next line)... dy/dx = 9/2 (sqrt x) - 6/(5 * y^2/5) * dy/dx. I have boldfaced the part which I believe is wrong; shouldn't it be dy/dx = 9/2 (1/sqrt x) - 6/(5 * y^2/5) * dy/dx? In other words, isn't the derivative of 9x^1/2 = 9/2 (1/sqrt x), instead of 9/2 (sqrt x), as he shows? Or am I missing something, here? At any rate, I would really appreciate it, if somebody who knows what they're doing would show me how to do the problem. I hope all of this was clear enough.

 

[DrPhil]

Okay, I am trying to work out a problem (problem # 16.5) found at the bottom of page 249 of Michael Kelley's book, The Humongous Book of Calculus Problems, and I just can't see how he got the answer that he did. After going over it for waaay longer than I care to admit, I think that I found an error in his solution, but then who am I, an old guy trying to learn calculus on his own, to tell? Here's the problem: "Given y = 9x^1/2 - 2y^3/5, find dy/dx." On the next page, p.250, he shows how he solves the problem, but it seems to me that he is wrong in his very first step, where he says, "Differentiate each term with respect to x," and then proceeds as follows: dy/dx = 9*1/2 (x^-1/2) - 2*3/5 (y^-2/5) * dy/dx (no problem, so far)...which all equals (next line)... dy/dx = 9/2 (sqrt x) - 6/(5 * y^2/5) * dy/dx. I have boldfaced the part which I believe is wrong; shouldn't it be dy/dx = 9/2 (1/sqrt x) - 6/(5 * y^2/5) * dy/dx? In other words, isn't the derivative of 9x^1/2 = 9/2 (1/sqrt x), instead of 9/2 (sqrt x), as he shows? Or am I missing something, here? At any rate, I would really appreciate it, if somebody who knows what they're doing would show me how to do the problem. I hope all of this was clear enough.

It is difficult to type algebra inline. For instance to make the order of operations unambiguous ..
for instance, (9/2)(sqrt(x)) as opposed to 9/(2 sqrt(x)).
You can't have too many parentheses! With a little effort, we can use the LaTeX formatting tool.

\(\displaystyle \displaystyle y = 9\ x^{1/2} - 2\ y^{3/5} \)

\(\displaystyle \displaystyle \dfrac {dy}{dx} = 9 \left( \dfrac{1}{2} \right) x^{-1/2} - 2 \left( \dfrac{3}{5}\right) y^{-2/5} \dfrac{dy}{dx}\)

\(\displaystyle \displaystyle \dfrac {dy}{dx} = \dfrac{9}{2\ \sqrt{x}} - \dfrac{6}{5\ y^{2/5}} \dfrac{dy}{dx}\)

After a couple more steps,

\(\displaystyle \displaystyle \dfrac {dy}{dx} = \left[ \dfrac{9}{2\ \sqrt{x}}\right] \div \left[ 1 + \dfrac{6}{5\ y^{2/5}} \right]\)

 

[rayroshi]

It is difficult to type algebra inline. For instance to make the order of operations unambiguous ..
for instance, (9/2)(sqrt(x)) as opposed to 9/(2 sqrt(x)).
You can't have too many parentheses! With a little effort, we can use the LaTeX formatting tool.

\(\displaystyle \displaystyle y = 9\ x^{1/2} - 2\ y^{3/5} \)

\(\displaystyle \displaystyle \dfrac {dy}{dx} = 9 \left( \dfrac{1}{2} \right) x^{-1/2} - 2 \left( \dfrac{3}{5}\right) y^{-2/5} \dfrac{dy}{dx}\)

\(\displaystyle \displaystyle \dfrac {dy}{dx} = \dfrac{9}{2\ \sqrt{x}} - \dfrac{6}{5\ y^{2/5}} \dfrac{dy}{dx}\)

After a couple more steps,

\(\displaystyle \displaystyle \dfrac {dy}{dx} = \left[ \dfrac{9}{2\ \sqrt{x}}\right] \div \left[ 1 + \dfrac{6}{5\ y^{2/5}} \right]\)

Okay, thanks for your time and trouble in making that response, Dr. Phil; I appreciate it. I guess that I was right, because I got the same answer that you did. I probably spent two to three hours, trying to see where I went wrong, when comparing my answer to the author's. But then I guess it'd be pretty hard to write a calculus book that has over 1,000 problems, without making a few mistakes, so I don't hold any hard feelings toward Kelley (the author). At this point in the game, I'm happy just to get a single answer correct, lol! I was interested by your comment about LaTeX. I did a little research on it, via Google, after reading your comment, but what I found was couched in vernacular terms that seemed to be meant for programmers, so I wasn't really able to glean much from it. Is there any particular source that I should go to that's better'n the others, as far as a "nuts and bolts"/how to manual, for learning it? Is it a program that makes it easy to format/write math formulas?

 

[DrPhil]

I was interested by your comment about LaTeX. I did a little research on it, via Google, after reading your comment, but what I found was couched in vernacular terms that seemed to be meant for programmers, so I wasn't really able to glean much from it. Is there any particular source that I should go to that's better'n the others, as far as a "nuts and bolts"/how to manual, for learning it? Is it a program that makes it easy to format/write math formulas?

"Easy" is a relative term. To see examples from posted equations, right click on the line, then choose "Show Math As -> TeX Commands." Whenever I see someone express something I haven't learned yet, I do that to see what the TeX commands are.

For a compendium of symbols and operators, I use a listing from
http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols

In this Forum, you need to put [FONT=MathJax_Main]\(\displaystyle [/FONT] in front of your LaTeX code and put [FONT=MathJax_Main]\)[/FONT] after your LaTeX code.

Enjoy - and Good Luck!