# Thread: equation of the tangent line?

1. ## equation of the tangent line?

the question is
find the equation of the tangent line to the curve 2x^3 + 2y^3 = 9xy when x=1, y=2

i tried
d2x^3/dx + d2y^3/dx = d9xy/dx

d2x^3/dx = 6x^2
d2y^3/dx = 6y^2
d9xy/dx = ?

i don't know how to do
how do i do this problem?

2. Originally Posted by haebinpark
the question is
find the equation of the tangent line to the curve 2x^3 + 2y^3 = 9xy when x=1, y=2

i tried
d2x^3/dx + d2y^3/dx = d9xy/dx

d2x^3/dx = 6x^2
d2y^3/dx = 6y^2
d9xy/dx = ?

i don't know how to do
how do i do this problem?
Use implicit differentiation. The RHS will require the product rule.

$\displaystyle \frac{d}{dx}(2x^3+2y^3) = \frac{d}{dx}(9xy)$

Looking at the RHS:

$\displaystyle u = 9x \: \rightarrow \: u' = 9$

$\displaystyle v = y \: \rightarrow \: v' = \frac{dy}{dx}$

$\displaystyle \frac{d}{dx}(9xy) = 9y + 9x \frac{dy}{dx}$

The LHS is simpler:

$\displaystyle \frac{d}{dx}(2x^3+2y^3) = 6x^2 + 6y^2 \frac{dy}{dx}$

Set both sides equal to each other and isolate dy/dx. You can then sub in the value you know to find the gradient of the tangent

Spoiler:
$\displaystyle 6x^2 + 6y^2 \frac{dy}{dx} = 9y + 9x \frac{dy}{dx}$

$\displaystyle 9x \frac{dy}{dx} - 6y^2 \frac{dy}{dx} = 9y - 6x^2$

$\displaystyle \frac{dy}{dx}(9x-6y^2) = 9y - 6x^2$

A factor of 3 will also cancel out: $\displaystyle \frac{dy}{dx} = \frac{3y-2x^2}{3x-2y^2}$

3. but
you have dy/dx on both side?

4. anyone...?

5. Originally Posted by e^(i*pi)
Use implicit differentiation. The RHS will require the product rule.

$\displaystyle \frac{d}{dx}(2x^3+2y^3) = \frac{d}{dx}(9xy)$

Looking at the RHS:

$\displaystyle u = 9x \: \rightarrow \: u' = 9$

$\displaystyle v = y \: \rightarrow \: v' = \frac{dy}{dx}$

$\displaystyle \frac{d}{dx}(9xy) = 9y + 9x \frac{dy}{dx}$

The LHS is simpler:

$\displaystyle \frac{d}{dx}(2x^3+2y^3) = 6x^2 + 6y^2 \frac{dy}{dx}$

Set both sides equal to each other and isolate dy/dx. You can then sub in the value you know to find the gradient of the tangent

Spoiler:
$\displaystyle 6x^2 + 6y^2 \frac{dy}{dx} = 9y + 9x \frac{dy}{dx}$

$\displaystyle 9x \frac{dy}{dx} - 6y^2 \frac{dy}{dx} = 9y - 6x^2$

$\displaystyle \frac{dy}{dx}(9x-6y^2) = 9y - 6x^2$

A factor of 3 will also cancel out: $\displaystyle \frac{dy}{dx} = \frac{3y-2x^2}{3x-2y^2}$

can i go

from 6x^2 + 6y^2 dy/dx = 9y+9x dy/dx

to 6y^2 dy/dx - 9x dy/dx = 9y - 6x^2 ?

and after finding dy/dx
how do i find the equation?
because that was the question... hmm