Question

- Can someone explain this differentiation concept to me please?

Answer 1

Lets say we are differentiating x^2 + xy^5 - 6x^3y + y^4

How do you differentiate with respect to just X or just Y?

For example, if you do with respect to X, does Y just become a constant?

Answer 2

it depends on the nature of the problem, there is implicit as well as partial differentiations

Answer 3

My apologies, this is implicit.

Answer 4

is it a partial differentiation if so yes u treat y as constant

Answer 5

if we assume that y is a function of x, then we use the chain rule. if we assume that x and y are independant variables ... then then are considered to be constants relative to each other

Answer 6

So can I ask a question?

Answer 7

you can

Answer 8

Lets say we are asked to find the equation of the tangent line to curve x^2 + 3xy^2 + y^3 = 5

Answer 9

x^2 +3x.[Y(x)]^2 +[Y(x)]^3 = 5 is correct ?

Answer 10

forget that the variables are named x and y, run thru the power rule, product rule, and constant rule as you would normally apply them.

Answer 11

and yes, that is a proper interpretation

Answer 12

My lecturer then writes : 2x + 3x.(2y.dy/dx) + y^2(3) + 3y^2.dy/dx = 0

Answer 13

dy/dx clutters things up, just denote it a y'

Answer 14

How does y^2(3) come ?

Answer 15

y^3 power rule, drop the exponent and subtract one

\[\frac d{dx}y^3=\frac {d(y)}{dx}3y^2=3y^2~y'\]

Answer 16

or, maybe that is attached to the product rule ... let me try that reasoning since i prolly looked at the wrong one

Answer 17

\[\frac d{dx}3xy^2=\frac {d(x)}{dx}3y^2+\frac {d(y)}{dx}6xy=3y^2+6xy~y'\]

Answer 18

yeah, product rule

Answer 19

uv' + vu'

3x = u
(2y.y') = v'
y^2(3) = v
3y^2 = u'

?

Answer 20

u = 3x
u' = 3

v = y^2
v' = 2y y'

Answer 21

2x + 3x.(2y.y') + y^2(3) + 3y^2.y'

Answer 22

I'm still a little confused where the last 3y^2 comes from?

Answer 23

y^3 ,power rule

Answer 24

[y(x)]^2 = y^2(3)

Answer 25

So x becomes 3?

Answer 26

x doesnt "become anything.

lets work this as a chain rule

\[[y(x)]^3=3[y(x)]^2~*~y'(x)\]

Answer 27

the (x) parts are implied therefore

y^3 = 3y^2 y'

Answer 28

implied = constant? (zero)?

Answer 29

implied simply means that the value of y is dependant upon the value of x by some rule that cannot be easily algebrated out

Answer 30

to imply that y is a function of x is to assume y = y(x)

Answer 31

if y was not dependant upon x for its value, then y and x would have no relationship and be constant relative to each other.


it is implied that when i close the door on the refridgerator, that the light turns off. The state of the light is implied by the state of the door.

Should we imply that when I turn on the water faucet that the light in the refridgerator turns on or off? no, then the state of the light is not implied by the state of the water faucet and any change in the rates have no effect on the rate of change in the other

Answer 32

I understand! Amazing explanation :)

Answer 33

Not the first time you've helped me amistre, thank you so much

Answer 34

:)