Question

can someone teach me implicit and explicit differentiation like I'm 5?

Answer 1

What is it that you don't understand?

Answer 2

Core concepts. My book doesn't explain it very clearly. For my first problem I'm supposed to find the derivative by implicit differentiation.

Answer 3

http://puu.sh/5cBFL.png

Answer 4

Well, you're given\[x^2y+xy^2=-2.\]Assuming that \(y\) is a function of \(x\), we differentiate implicitly as follows:\[2xy+x^2y'+y^2+2xyy'=0.\]Now it's all algebra; solve for \(y'\):\[y'=-\frac{y^2+2xy}{x^2+2xy}.\]Note that this is a differential equation.

Answer 5

Implicit differentiation of the first term
\[ \frac{d}{dx}(x^2y) \]
we use the product rule d(uv)= u dv + v du
\[ \frac{d}{dx}(x^2y)= x^2 \frac{d}{dx}y + y \frac{d}{dx} x^2 \\
x^2 \frac{dy}{dx} + y \ 2x \frac{d}{dx}x\\
x^2 \frac{dy}{dx} + 2xy
\]

Answer 6

Sam, do you understand why the \(\Large \dfrac{dy}{dx}\)'s are showing up in the problem?

Answer 7

previously I've written differentiation as a prime, like Y' or f'(x), so not really zepdrix

Answer 8

Primes are fine :) Just so long as you know where they're coming from.
I'm gonna ramble on for a minute here lol.
You don't have to read it if you don't want XD
I just wanna try to simplify things a tad if I can.

Take for example:\[\Large y=x^2\]You're probably used to simply writing the derivative operator with the prime on the left side of the equation. But what's actually happening is, we're taking the derivative of each side.\[\Large (y)'\quad=\quad (x^2)'\qquad\to\qquad y'\quad=2x\]

The tricky part with implicit differentiation is remembering that stupid chain rule!! :P
In the above example we could have applied the chain rule to the x^2, giving us:\[\Large y\quad=\quad x^2\qquad\to\qquad y'\quad=\quad 2x(x)'\qquad\to\qquad y'\quad=\quad 2x(1)\]But as you can see, the x' is insignificant.
The derivative of x with respect to x is just 1.

Answer 9

So if we take an expression like:\[\Large \left(xy\right)'\]Applying the product rule (similar to what phi showed you),\[\Large =\quad (x)'y+x(y)'\]But as we mentioned before, x' is simply 1,\[\Large =\quad (1)y+xy'\]

Answer 10

With implicit differentiation, when you take the derivative, you'll get a bunch of y' showing up in different places.

We need to group them up together so we can solve for y'.

Answer 11

These guys already gave some nice examples :O
You can let me know if you need to see another though.

Answer 12

Phi can you elaborate those steps??

Answer 13

which steps ?

Answer 14

How about
\[ \frac{d}{dx} x \]
?
you know that the answer is 1. one way to write this is
\[ \frac{d}{dx} x= \frac{dx}{dx} = 1 \]

if instead we did a constant, we get
\[ \frac{d}{dx} 2 = \frac{d\ 2}{dx} = 0 \]
if that makes any sense.

if y is a variable and depends on x we would expect that as x changes, y changes, and therefore the ration
\[ \frac{dy}{dx} \] exists and means something

Answer 15

*ratio

Answer 16

I understand how to find basic derivatives, I'm just confused in the dy/dx

Answer 17

what is
\[ \frac{d}{dx} y^2 \]
we use the chain rule d u^n = n u^(n-1) du
\[ \frac{d}{dx} y^2 = 2 y \ \frac{d}{dx} y= 2y \frac{dy}{dx} = 2 y \ y'\]

Answer 18

notice the difference between d/dx y^2 and d/dx x^2
\[ \frac{d}{dx} x^2 = 2x \frac{d}{dx} x = 2x \frac{dx}{dx}= 2x\]

Answer 19

can you do
\[ \frac{d}{dx} xy \]?

Answer 20

I see the difference but I don't know how to apply it

Answer 21

can you try d/dx (x y) ?

Answer 22

Khan's video might help
http://www.khanacademy.org/math/calculus/differential-calculus/implicit_differentiation/v/implicit-differentiation-1