|
|
|
United States24698 Posts
I think it would really help if you chunked/formatted and edited this O.o
|
|
|
|
|
For implicit differentiation, when you differentiate a term with both x and y in it, you need to use the product rule, i.e.:
(xy)' = (x')y + x(y') = y + x (dy/dx)
|
On May 31 2008 11:05 imDerek wrote:
For implicit differentiation, when you differentiate a term with both x and y in it, you need to use the product rule, i.e.:
(xy)' = (x')y + x(y') = y + x (dy/dx)
so..
xy + y^2 = 1 <-- all y = dy/dx as the deriv right?
x(dy/dx) + 2y(dy/dx) = 0 <-- am i done? im not sure if i did it right.
is wrong?
|
yup that's wrong.
The answer should be dy/dx = -y/(x+2y)
|
f(x) = (cos3)x^2 = (cos3)(2x) = 2xcos3 i thought cos would be -sin?
no idea what this is. cos^3(x^2)?
"f(x) = (x^3+1)^(-4/3) I know that the (-4/3) have to minus one and that would be (-7/3) the finalized result is f'(x) = -4x^2 (x+1)^(-7/3). I also understand how the -4 comes from (x^3 * (-4/3)) but not exactly know why the (x^+1) = (x-1) "
i don't know if that's what you have
"f(x) = cos(1-2x) --> -sin(1+2x) --> ? im stuck now. :/"
derivative of cos(BLAH) = -sin(BLAH)*derivative of blah = -sin(1-2x) * -2
"xy + y^2 = 1 <-- all y = dy/dx as the deriv right?
x(dy/dx) + 2y(dy/dx) = 0 <-- am i done? im not sure if i did it right."
you have to do a product rule for the first part. x*dy/dx+y+2y*dy/dx=0. LOL i see that you did that just forgot the product rule.
"
x^2 + xy + y^2 - 5x = 2
2x + x(dy/dx) + 2y(dy/dx) - 5 = 0
(whats the next step if i did it correctly?)"
again, product rule. the next step is to separate the dy/dx's and the non dy/dx's, so it'd be like
2x + x(dy/dx) + y + 2y(dy/dx) - 5 = 0
x(dy/dx) + 2y(dy/dx) = 5 - y - 2x
(x+2y)(dy/dx) = 5 - y - 2x
dy/dx=(5 - y - 2x)/(x + 2y)
"y^2 = ( (5x^2 + 2x)^(3/2) ) / 3
(this is ridiculous, what do i do first? it looks like a quotient rule but im not very sure.)"
Not a quotient rule since the 3 is on the bottom. Just do it normally implicitly..
"
x^2 + 2y^2 = 9
2x + 4y(dy/dx) = 0? "
yes right track
|
for this.
i have the answer sheet too, it only shows as this, which is why i am asking:
it doesnt have what you have, is there why? the answer sheet is from my prof.
--
f(x) = (cos3)x^2 = (cos3)(2x) = 2xcos3 i thought cos would be -sin?
no idea what this is. cos^3(x^2)?
|
are you sure there's no ^3 over the x? because besides that, it's the same thing
uhhh, what does cos3 mean lol. unless you mean like cos(3x^2) or like cos(3x)^2 or like cos^2(3x)
|
|
|
On May 31 2008 11:44 Rayzorblade wrote:
what
the
fuck
is
this
lol?
calculus. youll enjoy it when youre in it.
On May 31 2008 11:39 ydg wrote:
are you sure there's no ^3 over the x? because besides that, it's the same thing
uhhh, what does cos3 mean lol. unless you mean like cos(3x^2) or like cos(3x)^2 or like cos^2(3x)
oops, sorry, mistyped. the answer is
f'(x) = -4x^2 (x^3+1) ^ (-7/3) <-- this is the same as the answer? i mean, how?
as for the cos3, it starts off as this:
y = (cos 3)x^2 -> y' = (cos 3)(2x) = 2x cos 3. idk what it means, i think you know better than i. so which way should i look at this chain rule?
|
rofl. cos3 is a real number. Maybe it's clearer if instead of cos3 you write the number -0.989992496 (for 3 in radians).
|
oh it's the same, since the 3 in 3x^2 cancels with the /3 in 4/3 lol.
ohhh it's a number, gotcha. so it's just a constant times x^2, so it's like 3x^2 -> derivative would be 6x
so it's cos 3 times x^2 => 2 times cos 3 times x
|
Wtf u sure ur from China! (jk jk xD)
I'll do one:
xy + y^2 = 1
so take derivitive on both side, bear in mind that both x, y, are functions.
xdy + ydx + 2ydy = 0
(x+2y)dy = -ydx
dy/dx = -y/(x+2y)
|
when you've finished those, try to differentiate x^x
|
Hint: x^x=e^(xlnx).
lol actually, that gives it away.
|
try to integrate x^x w.r.t x instead
|
f(x) = (cos3)x^2 = (cos3)(2x) = 2xcos3 i thought cos would be -sin?
(x^2)cos(3), right?
cos(3) is a constant
f(x) = (x^3+1)^(-4/3)
[a(x^u)]' = au'[x^(u-1)]
+1 is a constant
Thus:
[(x^3+1)^(-4/3)]'
[(-4/3) (3x^2) (x^3+1)^(-7/3)]
[(-4x^2) (x^3+1)^(-7/3)]
f(x) = cos(1-2x)
(cos[u])' = -u'sin(u)
Thus:
[cos(1-2x)]'
(-2)sin(1-2x)
IMPLICITS <- I used to hate these :D!
xy + y^2 = 1 <-- all y = dy/dx as the deriv right?
x(dy/dx) + 2y(dy/dx) = 0 <-- am i done? im not sure if i did it right.
(uv)' = u'v + uv'
Thus:
[xy+y^2 = 1]'
[(1)y+x(y') + 2y = 0]
Isolate the y':
[y+2y = x(y')]
[(y+2y)/x = y']
[y' = (y+2y)/x] It looks better flipped around :[
x^2 + xy + y^2 - 5x = 2
2x + x(dy/dx) + 2y(dy/dx) - 5 = 0
(whats the next step if i did it correctly?)
(uv)' = u'v + uv'
[x^2 + xy + y^2 - 5x = 2]'
[2x + (1)y + x(y') + 2y - 5 = 0]
Isolate the y':
[2x+3y-5 = x(y')]
[(2x+3y-5)/x = y']
[y' = (2x+3y-5)/x]
y^2 = ( (5x^2 + 2x)^(3/2) ) / 3
(this is ridiculous, what do i do first? it looks like a quotient rule but im not very sure.)
Chain rule (I suck at this; might be wrong T_T):
[y^2 = ( (5x^2 + 2x)^(3/2) ) / 3]'
[2y(y') = (3/2) * (5x^2+2x)^(1/2) * (10x + 2) / 3]
[2y(y') = (5x+1) (5x^2+2x)^(1/2)]
[y' = [(5x+1) (5x^2+2x)^(1/2)] / 2y]
x^2 + 2y^2 = 9
2x + 4y(dy/dx) = 0?
(am i in the right track at least?)
Yes; isolate y -> ezgg
|
|
|
|
|
|
EPT
![[image loading]](http://www.texify.com/img/%5CLARGE%5C%21f%28x%29%20%3D%20%28x%5E3%2B1%29%5E%7B%28-4%2F3%29%7D.gif)
![[image loading]](http://www.texify.com/img/%5CLARGE%5C%21f%27%28x%29%20%3D%20%28-4%2F3%29%2A%28x%5E3%2B1%29%5E%7B%28-7%2F3%29%7D%2A%283x%5E2%29.gif)
![[image loading]](http://www.texify.com/img/%5CLARGE%5C%21y%5E2%20%3D%20%5Cfrac%7B%285x%5E2%2B2x%29%5E%7B%283%2F2%29%7D%7D%7B3%7D.gif)
![[image loading]](http://www.texify.com/img/%5CLARGE%5C%212y%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%20%5Cfrac%7B3%7D%7B2%7D%5Cfrac%7B%285x%5E2%2B2x%29%5E%7B%281%2F2%29%7D%7D%7B3%7D%20%2A%20%2810x%20%2B%202%29.gif)
![[image loading]](http://www.texify.com/img/%5CLARGE%5C%21-4x%5E2%28x%2B1%29%20%5E%20%28-7%2F3%29.gif)
![[image loading]](http://www.texify.com/img/%5CLARGE%5C%21f%28x%29%20%3D%20%28cos3%29x%5E2.gif)
