Question

when finding the dirivitive of f(x,y) =2x^2y^3
using the fuction at (a,b) and holding y constant.. you get g(x)=f(x,b) =2x^2b^3 ....
THEN rate of change of the function at (a,b) if we hold y fixed and allow x to vary... g'(a)=4ab^4 ....... where does this part come from, how do you get to 4ab^4 from 2x^2b^3?

Answer 1

Hmm shouldn't you get 4ab^3? Is that a typo by chance?
Maybe I'm mistaken :) Lemme look at the problem again.. hmm

Answer 2

Hmm I guess this one has me stumped. :\
@Hero @hartnn @experimentX

Answer 3

Holding 'y' constant means partially differentiating w.r.t 'x'
\(\large \frac{\partial}{\partial x}(2x^2y^3)=4xy^3 \\then, g(x)=f'(x,b)=4xb^3\)
in the question it appears that derivative is not taken at all, even if its mentioned..

Answer 4

i would request the asker to post the screenshot of actual question.

Answer 5

this is what i am looking at http://tutorial.math.lamar.edu/Classes/CalcIII/PartialDerivatives.aspx where does the 4ab(cubed) come from?

Answer 6

So it was a typo? Darn I wish you woulda mentioned that :) lol

Answer 7

right, so since 'y' was constant, it was taken as b, (y=b)
hence, the function f is now only function of x = g(x) = \(2x^2b^3\)
Now, to get rate of change, we differentiate w.r.t x
\(g'(x)=4xb^3\)
to find rate of cahnge AT x=a
we just plug in x=a in that.
\(g'(a)=4ab^3\)
got this ?

Answer 8

another way to represent g'(a)
\(g'(a)=4ab^3=f_x(a,b)\)

Answer 9

I am having trouble doing anything on this site, it is lagging like hell.. have a few questions regarding partial derivitives.. if you guys can shoot me an email that would be awesome.. fivestarsmi@gmail.com

Answer 10

everything you post comes up as math processing error, I cant even view it =/

Answer 11

let me post you an image...of my comment...

can u view this ? and understand ?

Answer 12

I understand everything, except where the 4xb^3 comes from

Answer 13

how do you get from 2x^2b^3 to 4xb^3?

Answer 14

the derivative part ??
derivative of x^n is nx^(n-1)
so, derivative of 2x^2 will be 2*(2x^{2-1}) = 4x
and since b^3 is constant, it can be taken out of derivative.
so,
(2x^2b^3)' = 4xb^3

Answer 15

(2x^2b^3)' =2b^3(x^2)'=2b^3(2x)= 4xb^3

Answer 16

any further doubts ?

Answer 17

is there a name for this formula to find the dirivitive? the derivative of x^n is nx^(n-1)

Answer 18

is there a name for this formula to find the dirivitive? the derivative of x^n is nx^(n-1)

Answer 19

no... i don't think there is any name to it...it helps in differentiating any polynomial....

Answer 20

how do you describe differntiating in laymans terms lol

Answer 21

if a function f(x) is increased by very small amount say, 'h' (small means h->0)
then differentiation of 'x' or rate of change of the function f(x) = f'(x)
will be just the ratio of change in function / small change h.
and the mathematical formula will be
lim h->0 {f(x+h)-f(x)}/{h}
i don't know whether this is layman's terms, but ask if you have any doubts in this....

Answer 22

ok, in your first reply you have the partial differential symbol of the partcial dif of x... does that just mean to apply the x^n is nx^(n-1) formula? ... how is that read?

Answer 23

the question had x^2 in it, thats why i used x^n formula.
i read it as 'daaba by daaba x' but different people read it differently.
technically its partial derivative of function f w.r.t x

Answer 24

qhat dould it mean if it had daaba z by dadda x

Answer 25

for instance in the navier-stokes formula.. it has daaba v by dabba t ... is the v for viscosity or vector?

Answer 26

v is for flow velocity..

Answer 27

z can be function of many variable, like x,y,p,q....and so on
now partial derivative of z w.r.t x means
treat all the variables other than 'x' as CONSTANT.
now z will only be function of 'x' and just diff. w.r.t x normally.

Answer 28

so daaba v by daaba t .. the partial dif of the flow velocity over the partial dif of t? (time?)

Answer 29

the partial dif of the flow velocity with respect to 't'
means only 't' is variable, all other letters are constant

Answer 30

Cartesian coordinates

Writing the vector equation explicitly.... what does this mean, when you have several partial dirivities in an equation/.

Answer 31

ok, 'u' is the function of t,x,y,z
daaba u by daaba t implies derivative of 'u' treating x,y,z as constants and 't' as variable.
daaba u by daaba x implies derivative of 'u' treating t,y,z as constants and 'x' as variable.
daaba u by daaba y implies derivative of 'u' treating x,t,z as constants and 'y' as variable.
daaba u by daaba z implies derivative of 'u' treating x,y,t as constants and 'z' as variable.

Answer 32

what does smooth mean?

Answer 33

smooth means not rapidly changing.....

Answer 34

s = (x, y, z) \[\in\] ℝ3 does \[\in\]just means in the 3d space?

Answer 35

∈ means 'belongs to'
R3 means 3 dimentional space.