Question

Use implicit differentiation to find the equation of the tangent line to the curve xy 3 +xy=6 at the point (3,1) .

The equation of this tangent line can be written in the form y=mx+b where m=? and b=?

Answer 1

is this your equation?

xy^3 + xy = 6

Answer 2

yes, sorry

Answer 3

x'.y^3 + 3xy^2.y' + x'.y + x.y' = 0 ; where x' = 1; solve for y'

Answer 4

\[y^3 + 3xy^2.y' + y + x.y' = 0\]
\[3xy^2.y' + x.y' = -y^3-y\]
\[y'(3xy^2 + x) = -y^3-y\]
\[y' = \frac{-y^3-y}{3xy^2 + x}\]

Answer 5

your point gives you your x and y value; so plug those in for your slope

Answer 6

\[y' = \frac{-1-1}{3(3)(1)+3}\]
\[y' = -\frac{2}{12}=-\frac{1}{6}\]

Answer 7

\[tangent=(-1/6)x -(-(3)/6) + 1\]
\[tangent=(-1/6)x +1/2 + 1\]
\[tangent=(-1/6)x +1.5\]

Answer 8

so with those you just redistribute and solve for y'?

Answer 9

yes

Answer 10

implicit diff is the same as explicit; in fact explicit is the case where you only have a single y term

Answer 11

which makes the latter easeir :/

Answer 12

\[\frac{d(y)}{dx}=\frac{dy}{dx}\]
\[\frac{d(y^2)}{dx}=\frac{dy}{dx}2y\]
\[\frac{d(xy)}{dx}=\frac{dx}{dx}y + x\frac{dy}{dx}\]

Answer 13

its not that its easier; its just that in case of an "implicit" the solution to dy/dx involves terms of y :) when youre used to there not beig any

Answer 14

the power rule is stillthe power rule; the quotitent rule is still the quotient rule; the chain rule is still the chain rule etc etc etc

Answer 15

yeah, but it also makes it that much harder to shortcut, using a calculator for example.

Answer 16

the only reason youre used to throwing out the derived bits; is becasue dx/dx = 1; but if an equation is derived with respect to an other variable like time:

\[x^2 + xy = 67yx^2\]
\[Dt(x^2 + xy = 67yx^2)\]
\[\frac{dx}{dt}2x + \frac{dx}{dt}y + x\frac{dy}{dt} = \frac{dx}{dt}134yx + 67x^2\frac{dy}{dt}\]

Answer 17

lol .... yeah, calcs hate multivariables