Question

Find the slope of the tangent line to the curve
3 x^2 - 2 xy + 1 y^3 = 72
at the point ( -3,3 ).

Answer 1

what is the "1" doing there?

Answer 2

its in the problem

Answer 3

thats how they wrote it...

Answer 4

with a "1" as in \(1\times y^3\) very odd indeed

Answer 5

yep

Answer 6

lol

Answer 7

start with
\[6x-2y-xy'+3y^2y'=0\] replace x by -1 and y by 3 and solve for \(y'\) or else solve for \(y'\) using algebra first and then make the replacement. no matter

Answer 8

ok lets see

Answer 9

damn typo
\[6x-2y-2xy'+3y^2y'=0\]

Answer 10

the only calculus part of the problem is finding the derivative. do you get that part?

Answer 11

i dont understand how u get the implicit differentiation....

Answer 12

ok think of it this way:
you are pretending y is a function of x, so think of y as \(y=f(x)\)

Answer 13

ok

Answer 14

so for example, the derivative of \(y^3\) is like taking the derivative of \(f^3(x)\) by the chain rule you get \(3f^2(x)f'(x)\) but you write \(3y^2y'\) which is more brief

Answer 15

ok i see

Answer 16

and for \(xy\) you treat it like \(xf(x)\) so you need the product rule, just as you would it if was \(x\sin(x)\)

Answer 17

so what if u have 3xy^2?

Answer 18

then you need both the product and the chain rule.
suppose you had \(3x\sin^2(x)\)

Answer 19

you would have to use the product rule, because it is a product, and you have to use the chain rule to find the derivative of \(\sin^2(x)\)

Answer 20

so if you have \(3xy^2\) you think of it as \(3xf^2(x)\) and the derivative would be
\[3(f^2(x)+x\times 2f(x)f'(x)\] by both the product and the chain rule
but you just write
\[3(y^2+2xy')\]

Answer 21

ok it makes sense.....thanks...

Answer 22

yw

Answer 23

actually there is a another more prosaic way of looking at this. got to the video on implicit diff here
http://justmathtutoring.com/
to get a different take on it if you need more

Answer 24

k thanks so much