Question

Could somebody help me to understand implicit differentiation?

Answer 1

Bring the equation.

Answer 2

I don't want to cheat, I just want to understand the concept.

Answer 3

For instance, how would you differntiate ax^2+byx+cy^2=d?

Answer 4

I don't give answers if you thought that's what I was gonna do. Lol

Answer 5

Implicit differentiation just makes no sense to me without units (like meters, ml/s, etc.)

Answer 6

Implicit differentiation is a special case of the chain for y function written implicitly in terms of x.

Answer 7

so you have the equation
y= x
assume that over time x changes. we write the change of x with respect to time as
\[ \frac{dx}{dt} \]

now y = x, so if x changes, y also changes.
it might be clear that in this case the change in will equal the change in x
\[ \frac{dy}{dt}= \frac{dx}{dt} \]

Answer 8

or we could start with
y= x
and ask, how does y change when x changes (the change in y with respect to x)
\[ \frac{dy}{dx} = \frac{dx}{dx} \\ \frac{dy}{dx} = 1\]

Answer 9

so far, does that make sense ?

Answer 10

Good explanation @phi

Answer 11

Yeah, phi, for the most part.

Answer 12

Sorry, I'm home-schooled and my teachers are already on break.

Answer 13

now say we start with
x+y =1

and ask how does this equation change with respect to x
we would take the derivative with respect to x of both sides
\[ \frac{d}{dx}(x+y) = \frac{d}{dx}1\\
\frac{d}{dx}x+\frac{d}{dx}y = 0 \\
\frac{dx}{dx}+\frac{dy}{dx}=0\\
1+\frac{dy}{dx}=0 \\
\frac{dy}{dx}= -1 \]
notice that even though we did not start with y alone on the left side, we were able to find dy/dx.

Answer 14

You also didn't move it over to the other side... For some reason, I was stuck in the mindset that the terms with y as a factor had to moved to the same side.

Answer 15

yes, you could start by solving for y. It is easy for the equation
x+y =1
we get
y = -x + 1
now take the derivative with respect to x of both sides
\[\frac{dy}{dx} = -1 \]
just like above.

but sometimes the equations are much uglier, and it is hard to solve for y... but you still want to find dy/dx
implicit differentiation is how we do it.

If you have time, you might want to watch
http://www.khanacademy.org/math/calculus/differential-calculus/implicit_differentiation/v/implicit-differentiation-1

Answer 16

I already watched khan academy... That's why I'm here;I'm still not understanding. Calculus makes me feel so stupid!

Answer 17

You gave me a good idea of what I'm supposed to be doing, though. Thank you.

Answer 18

Calculus uses new ideas from algebra or geometry or trig (but it also uses those subjects). So there are 2 ways to NOT understand it:
1) you don't remember or did not learn some of the geometry, (or algebra or trig)
2) you do not "get" the new ideas (yet)

Answer 19

I'm in the second category... and the first (I skipped trig because I'm an idiot and I thought I was SOOO smart that I could).

Answer 20

The main idea is that you can plot a curve
example x^2 + y^2 = 4
(this is a circle centered at (0,0) with a radius of 2)

and ask the question, at some point on this curve, what is the slope of the curve?
Of course, it makes sense to talk about the slope of a straight line... but a curve is not straight... so the new idea is that a curve can have a slope at a specific point.
(the slope of the curve at a point is the slope of the *tangent line* that just grazes the curve at that point)

Answer 21

Phi, how do you deal with terms that contain x and y?

Answer 22

*both x and y

Answer 23

we can try to find the slope of the curve
x^2 + y^2 = 4

First, some common sense:
1) there is no *one slope* (1 number like in a straight line) because the curve is not straight.

Not so common sense:
the slope will be a *single equation* , where you plug in the (x,y) of interest, and you are able to compute the slope.

now to business. the slope is change in y divided by change in x... in calculus dy/dx

take the derivative of this equation with respect to x:
\[ \frac{d}{dx}(x^2 + y^2 = 4 )\]

Answer 24

\[ \frac{d}{dx}x^2 + \frac{d}{dx}y^2 = \frac{d}{dx}4 \\
2 x \frac{d}{dx}x + 2y \frac{d}{dx}y = 0 \\
2x + 2y \frac{dy}{dx}=0
\]
solve for dy/dx to find the equation for the slope.

Answer 25

I meant like xy in x^2 +xy + y^2

Answer 26

First, can you do the x^2 + y^2 = 4 problem ?

Answer 27

dy/dx=-x/y

Answer 28

I know that part, I was tripped up by the part with xy

Answer 29

the ideas are the same for more complicated functions
\[ \frac{d}{dx}(x^2 +xy + y^2 = 1)\]
term by term what do you get ?

Answer 30

\[\frac{d}{dx} x^2 +\frac{d}{dx}( xy) +\frac{d}{dx} y^2 = \frac{d}{dx} 1 \]

Answer 31

I would assume that dy/dx=-2x/(2y+1)

Answer 32

Do you take the derivative with respect to x, y, or both on the xy term?

Answer 33

btw, notice with implicit differentiation we start with an equation

you should get
\[2x + \frac{d}{dx} (xy) + 2y \frac{dy}{dx} = 0\]

for the xy, we use the product rule
d(uv) = u dv + v du
\[ \frac{d}{dx} xy = x \frac{d}{dx} y + y \frac{d}{dx} x\]

Answer 34

WHY DIDN'T I THINK OF THAT?!?

Answer 35

THANK YOU SO MUCH!!! You just made my day! Now I can actually figure this stuff out!

Answer 36

we now get the slightly messy equation
\[2x + x \frac{dy}{dx} + y + 2y \frac{dy}{dx} = 0 \]
we use algebra to "solve" for dy/dx

Answer 37

So, \[dy/dx\] of \[3x ^{2}-2xy+4y ^{2}=0\] is \[(y-3x)/(x-4y)\]

Answer 38

I got (y-3x)/(4y-x)

Answer 39

Well, I was close.

Answer 40

Let me explain my logic...

Answer 41

here are the details (I'll use y' for dy/dx easier to type)
\[ 3x ^{2}-2xy+4y ^{2}=0 \\
6x -2 ( x y' + y) +8y\ y'=0 \\
3x -( x y' + y) +4y\ y'=0 \\
3x - x y' -y +4y \ y'=0\\
4y \ y'- x y'= y-3x\\
(4y -x) y'= y-3x\\
y'= \frac{y-3x}{4y-x}
\]

Answer 42

\[d/dx(f)=6x-2dy/dx(xy)+dy/dx(8y)=0\]
\[d/dx(xy)= xy'+x'y, x'=1, y'=dy/dx\]
\[d/dx(xy)=dx/dy(-2x)-2y\]
\[d/dx(f)=6x-2y+dy/dx(8y-2x)=0\]
\[dy/d(4y-x)=y-3\]
\[dy/dx=(y-3x)/(4y-x)\]
I made the oldest algebraic mistake in the book... I ADDED across instead of subtracting. Thanks for catching my mistake; I was worried for a second, then I worked it out again.

Answer 43

yes. I assumed you lost a sign somewhere in the first try. Not a basic misunderstanding... just a simple mistake.

Answer 44

Well, until you explained that I should use the product rule, I was completely lost.

Answer 45

Thanks again.

Answer 46

yw