Question

Implicit Differentiation

Answer 1

AHA you are my hero!

Answer 2

ok, so as far as where these functions are coming from, they are probably just making them up, so i wouldnt get too hung up on trying to figure that aspect out.

Answer 3

What matters is not what the function is, but how to actually calculate the derivative. Lets look at their first example:
\[x^2+y^2 = 25\]

Answer 4

see i think...well THOUGHT I got that part, I was moving along pretty well until I got down to the part where they start using letters rather than numbers and the section with a whole bunch of annoying single words on the left side of the page and whooseywhatsey in the center...i THINK I get most of it up to there :/

Answer 5

yeah i see what you are talking about, thats just a notation thing. When doing this, you want to differentiate everything like normal, except whenever you the the derivative of 'y', you tack on a y' (or dy/dx, whichever you like to use more). When you do this you are using the Chain Rule.

Answer 6

So with the example above, it would be:
\[x^2+y^2 = 25 \Longrightarrow 2x+2y(\frac{dy}{dx}) = 0\]

Answer 7

AHA! the chain rule is the source of all of the evil!

okay so for that what is the formula you would use to get from the first bit to the last? or not formula...the rule!

Answer 8

i dont think i understand the question >.< are you asking how i would finish the problem?

Answer 9

well how did you get from the part before the arrow to the part after?

Answer 10

ah, i differentiated. I normally dont show notation for that, but if i had to it would be something like:
\[x^2+y^2 = 25 \Longrightarrow \frac{d}{dx}(x^2+y^2) = \frac{d}{dx} 25 \Longrightarrow 2x+2y\left(\frac{dy}{dx}\right) = 0\]

Answer 11

\[\frac{d}{dx}\]
means im taking the derivative with respect to x.

Answer 12

hmmmmmmmm okay this is starting to get much less fuzzy...haha okay its two in the morning so one last question before i get some rest, why exactly does IMPLICITness exist? I'm hoping there's a reason for it that's just way over my head but it seems like you could just swap things around and turn it "explicit" and whatnot...oh math...

Answer 13

Normally when dealing with functions they are in the form:
\[y = f(x)\]
Where y is the only variable on one side and there are nothing but x's on the other side.
Now we are running across equations in the form of:
\[f(x,y) = c\]
Where we have x's and y's all mixed up with each other. Sometimes it might be easy to take that f(x,y) and solve for y to make it like the old equations, but sometimes its not.
Implicit Differentiation is a technique that works well with f(x,y) because you dont have to solve for y. You differentiate implicitly, and then solve for dy/dx.

Answer 14

hey i need ur help .......
can u differentiate y=sqr(x+1)
using the first principle of differentiation

Answer 15

is that by using the difference quotient?
\[\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}\]

Answer 16

yes

Answer 17

it is also called the 4 -step rule

Answer 18

hmm...im not really getting anywhere >.< im still tinkering away though lol

Answer 19

oh!!!!!! lol

Answer 20

oh, i was forgeting to put the 1 in >.< my bad:
\[\frac{\sqrt{x+h+1}-\sqrt{x+1}}{h} *\frac{\sqrt{x+h+1}+\sqrt{x+1}}{\sqrt{x+h+1}+\sqrt{x+1}}\]
\[= \frac{x+h+1-x-1}{h(\sqrt{x+h+1}+\sqrt{x+1})} = \frac{h}{h(\sqrt{x+h+1}+\sqrt{x+1})}\]
\[ = \frac{1}{\sqrt{x+h+1}+\sqrt{x+1}}\]
Now its ok to take the lim as h goes to 0:\[\lim_{h\rightarrow 0}\frac{1}{\sqrt{x+h+1}+\sqrt{x+1}} = \frac{1}{\sqrt{x+1}+\sqrt{x+1}} = \frac{1}{2\sqrt{x+1}}\]

Answer 21

thnx.....:)

Answer 22

our hero joe!!

Answer 23

What about: \[(7x-y+1)\tan(5x+xy^2-3)=1\]