taking derivatives of functions and their rules?

given the following functions:

z^2 = x^2 + y^2

What would be the proper way of differentiating it? My instincts tell me it should be 2z = 2x + 2y, but I've been told it's not always correct because of what respect you're deriving the function from or to.

Can anyone show me an easy way to remember all these derivative rules rather than just pure memorization? It's ok if it's only a few things I just have no choice but to memorize, but it seems as though my professor and this book want me to just "memorize" things rather than learn it and prove it if I ever needed to.

Thanks in advance to any/all replies.

P.S. Are there any other zones/sites where I can ask these kinds of questions and get back faster responses? The quality of responses in this zone so far are impeccable, but it's a bit frustrating when I have to wait an entire day for a solution to a problem and times isn't on my side. Not annoyed, just curious. Thanks again.
Who is Participating?

[Product update] Infrastructure Analysis Tool is now available with Business Accounts.Learn More

I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

Your equation does not define a function unless you restrict the domain; say, for example, for z non-negative. For a given value of x and y, there are 2 values of z. (For example, if x=3, and y=4, then z = +5 satisfies the relation.

Without restriction, the equation defines a relation. With the restriction that z is non-negative, then you could write:
   z = sqrt( x^2 + y^2 )

Now you have a function and you can take derivatives - i.e., partial derivatives. But, it's a lengthy discussion.
In these links you will see surfaces that represent the function z = f(x,y). If you were to pick a particular value of x, say x = 5, then what you have is a plane in the 3d universe where x = 5; and this plane cuts the surface to form a function z = g(y); and there is no x since it is just a constant. So when you take the ordinary derivative of z=g(y), you are actually taking the partial derivative z = f(x,y) with respect to y. That is, for a constant value of x, we vary y and see the effect on the change in z.


There is another way to look at this - Implicit Functions:

With the restricted branch that z is non-negative, then you say that z is a function of x and y; i.e., z = z(x,y).

    z(x,y) - sqrt( x^2 + y^2 ) = 0

Let F(x,y,z) = z(x,y) - sqrt( x^2 + y^2 )

Then your original formula is now F(x,y,z)  = 0, and you can take partials. Here are some interesting results that show relationships between different partial derivatives:
ZenotureAuthor Commented:
Ok, I was hoping to maybe break the question up into parts to understand it better, but that might've made things more difficult. So here is the whole question

if z^2 = x^2 + y^2, dx/dt = 2, and dy/dt = 3, find dz/dt when x = 5 and y = 12.

If I were to try and solve this, by merely looking at it, my first impulse would be to take the derivative of both sides giving me dz/dt = 2(5) + 2(12) = 34; z = sqrt(34)

But that's wrong, so I need a different way to see it or think of it so I don't get carried away on first impulse and lose points.
Your answers would be quicker if your question was clearer. Here you have a function z of what seems  like 2 variables x and y, ie

   z² = x² + y²

To me it looks as if you are doing partial derivatives, am I right? In which case you have a partial derivative wrt x and also wrt y

Differentiating both sides wrt x gives

  2z*(dz/dx) = 2x


dz/dx = x/z = x/sqrt( x² + y² )

and by similar analysis

dz/dy = y/z =  y/sqrt( x² + y² )

atm I don't quite understand your question.

Break your question into smaller pieces so we can answer more clearly. We are really good at this differentiating stuff here.

C++ 11 Fundamentals

This course will introduce you to C++ 11 and teach you about syntax fundamentals.

This new phrasing of your question is actually a new question. But maybe this will give you an idea:

z^2 = x^2 + y^2
z^2 = 5^2 + 12^2 = 25 + 144 = 169
So, z = +13; but let's just deal with +13

Implicit differentiation:
2z dz = 2x dx + 2y dy

You can divide by dt:
2   z  dz/dt = 2  x dx/dt + 2    y dy/dt
plug in known values. I got dz/dt = 92/26 but check the work to see if there's an arithmetic mistake.

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
Take a look at this. I suspect it will come in handy someday.
what you really have a curved line through 3d space.

t could be time, for example. So, for any value of t, you have a point (x,y,z).
But you weren't told what the above three functions were. You were just given an implicit relationship between x, y, and z from which we came up with:

Implicit differentiation:
2z dz = 2x dx + 2y dy
which means that at a give point (x,y,z), if you know the change in dx and dy, then you know the change in dz from this implicit differentiation result.
> What would be the proper way of differentiating it?
Differentiating what, with respect to what?
ZenotureAuthor Commented:
At first I was confused why you just decided to divide by dt, but it was just to show the thinking process, so that helped a lot. Thanks.
...so what on earth happened there?
Just for the record, as there are some confusing posts here,

Problem : if z^2 = x^2 + y^2, dx/dt = 2, and dy/dt = 3, find dz/dt when x = 5 and y = 12.

Solution :

Take the total differentials with respect to t :-

    2zdz/dt = 2xdx/dt + 2ydy/dt

simplify and eliminate z (because z is not a given value)

    dz/dt = (xdx/dt +ydy/dt)/sqrt(x²+y²)

subsitute the given values :-

    dz/dt = (5*2 +12*3)/sqrt(5²+12²) = (10+36)/sqrt(169) = 46/13 = 3.54 roughly

(which corresponds to phoffric's 92/26)

>> At first I was confused why you just decided to divide by dt
That was just a heuristic shortcut that you will often see. Since  x=x(t),    y=y(t), and   z=z(t), then if you want to take the derivative of some function, say w(t) = [ x(t) ]^2, with respect to t, then you can use the chain rule. Here is a short tutorial on the chain rule that has a nice step by step flash presentation for their examples:

So the derivative of w(t) with respect to time using the chain rule is done as:
     d/dt[ w(t) ] = d/dt{ [ x(t) ]^2 } = [ 2  x(t) ] d/dt[ x(t) ] = 2 x dx/dt

ZenotureAuthor Commented:
>> http://archives.math.utk.edu/visual.calculus/2/chain_rule.4/index.html

Thanks for the link, I was actually having trouble understanding chain rule, so that was a huge help. What I don't understand is why the book didn't explain chain rule so simply...

Additionally, It's hard to find research online, or better ways/methods of solving problems because (as states in another post) my professor likes things done in a certain way, and even if we get the right answer, we lose points for not following his method, but never does well in explaining it...
>> even if we get the right answer, we lose points for not following his method
Sounds like the course is interested in understanding the math and methodology. That part is good. On the flip side, if you get the wrong answer (say an arithmetic error), but show understanding of the math principles, then you should get partial credit.

>> but never does well in explaining it
This occurs more frequently in online courses. My wife took a course, where all the professor said was "you are right", or "you are wrong". In a classroom, you can ask questions during the lecture or catch the professor as he is walking out with your specific question.

An aeronautical engineering PhD candidate complained to me that his PhD thesis was rejected because he didn't follow the standard methodology; and that he got to the correct conclusions by taking shortcuts. His argument to them was that his shortcuts were deserving of praise for their originality and providing a new methodology that always works. He was offered another chance to complete the thesis provided he pay for another 1.5 years of tuition. He just told them to @#$ off, and now he has a mere Master's degree.
ZenotureAuthor Commented:
>> This occurs more frequently in online courses.
That's the thing, this is a face-to-face class, and I go to his office to get further clarification, which actually only just confuses me more =) (ironically, the whole point of taking this class in person was to avoid situations such as these)

>> An aeronautical engineering PhD candidate...
Sounds about right, and probably what I would do if it were me =P
If a solution works and can be proven, it should be valid, but not everyone sees that way, the game of life =)
There are levels of abstractions in academic subjects leading to a specific variety of shop talk. And these higher abstraction levels are necessary for progress towards more complex topics.

But when a student has trouble understanding the current level of shop talk, it would be nice if the professor understood how to drop from their lofty tower and help pull the student to a higher level. Sometimes, it is simply an inability for the professor to remember how to explain at a lower level; and they'll defer you to the TA's (who unfortunately may understand less than you).

On the other hand, it is sometimes by design that they remain obfuscated. As one physics professor explained to me while carrying his surf board to the beach, "I spent a lot of time as a student suffering to get where I am; and I'm not going to make it easier for others to get there." (Ok, I can't remember the exact word-for-word quote, but most of the words and spirit were included.) I was shocked by this sink-or-swim attitude from a professor. At work this attitude was often in place; but then again, the big difference is that we are the ones being paid.
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
Math / Science

From novice to tech pro — start learning today.