Question

I was given an example of how to derive a derivative from an implicit function, but I don't understand how they got their answer. (will post below)

Answer 1

\[y^4+3y-4x^3=5x+1\]
The answer was \[y^4y'+3y'-12x^2=5\]

Answer 2

The one thing that confuses me is the y's. Are you supposed to add another y variable?

Answer 3

they are taking the derivative with respect to x
use the chain rule
\[ \frac{d}{dx} \left( y^4+3y-4x^3=5x+1 \right) \]
\[ \frac{d}{dx} y^4 + 3 \frac{d}{dx} y -4 \frac{d}{dx} x^3 = 5 \frac{d}{dx} x + \frac{d}{dx} 1\]

Answer 4

\[ \frac{d}{dx} y^4= 4 y^3 \frac{d}{dx} y \\
4y^3 \frac{dy}{dx} \]

Answer 5

This is the chain rule. Pretend y (which is a function of x) is f(x). Then whenever you see something like:

\[(y^7)' = 7y^6 y'\]
This is the same as:
\[((f(x))^7)' = 7(f(x))^6*f'(x)\]

Answer 6

another example
\[ -4 \frac{d}{dx} x^3= -12 x^2 \frac{dx}{dx} \\ =-12x^2 \]

Answer 7

Well your other example with 'x' makes sense, but are you supposed to just use the power rule for x and us the chain rule for a y variable?

Answer 8

you always use the chain rule. But when you take the derivative with respect to x you get a dx/dx that "goes away" (see example with x^3)

Answer 9

Is there a reason why it goes away for x and not y?

Answer 10

another example:
\[ \frac{d}{dx} (x+1)^2 = 2(x+1) \frac{d}{dx} (x+1) \\
= 2(x+1) \left(\frac{d}{dx} x+\frac{d}{dx} 1 \right)
\]
and that becomes
\[ 2(x+1) (1 +0) = 2(x+1) \]

Answer 11

I think of it like this:
\[ \frac{d}{dx} x = \frac{dx}{dx} = 1\\ \frac{d}{dx} y= \frac{dy}{dx} \]

Answer 12

Kind of unrelated but amusing, if you treat x as a function and differentiate it with the chain rule, you will get an infinite 'chain':
\[x' = 1*x^0 x'=1*x^0*1*x^0*x' =1*x^0*1*x^0*1*x^0*x' \]
haha I was hoping this would give some clever way of proving x'=1 but I don't think we can do that here, oh well.

Answer 13

That's interesting to think about.

Okay here's an example. Tell me if I'm right...

Answer 14

\[2y^7 \rightarrow 14y^6y'\] and \[2x^3 \rightarrow 6x^2\]

Answer 15

yes

Answer 16

.

Answer 17

Sorry guys. Openstudy just crashed and reloaded itself randomly. Weird?

Anyways, just to make sure, is the "extra y" always added on when taking the derivative of a y variable?

Answer 18

if we take the derivative with respect to a third variable, such as t we would do
\[ \frac{d}{dt} (x^2 + y^2) = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \]

notice the "prime notation" y' leaves out information (we have to remember that it means dy/dx)

also, you sometimes see Newton's notation, where we put a dot over the variable to denote the derivative with respect to t (time)

\[ \frac{d}{dt} (x^2 + y^2) = 2x \dot{x} + 2y \dot{y} \]

Answer 19

I am sorry, but I am completely lost.

Answer 20

***Is there a reason why it goes away for x and not y?***

you write the ratio of two "infinitesimals" dy/dx
if the top and bottom are both dx as in dx/dx that ratio is 1

Answer 21

***I am sorry, but I am completely lost.***

do you know how to use the chain rule?

Answer 22

Yes, but the y and x dilemma was something never explained to me.

Answer 23

Oh! I think I'm starting to understand now!

Answer 24

so because it's dy/dx it would keep the y variable
and like you said dx/dx = 1 so the x would just turn into 1.

Answer 25

yes

Answer 26

Finally I get it! :D

And it would do that with any variable right?
dt/dx = keep the t
da/dx= keep the a

Answer 27

yes

Answer 28

I sorry. At times it takes me a minute to catch on.
Thank you so much for your help.
As well as Empty.

Answer 29

yw