# Thread: Equation of a tangent line

1. ## Equation of a tangent line

Okay I know I am doing something wrong but I can't see my mistake - please help me

"Find the equation of the tangent line to the curve
3x^2 - 3y^3+2yx^2=-45 at point (-2,3)

So first my goal is to find the derivative fox x so here I go....

6x - 9y^2 (dy/dx) + 4xy + 2x^2(dy/dx) = 0

dy/dx (-9y^2 + 2x^2) = -6x -4xy

dy/dx = (-6x-4xy)/(-9y^2 + 2x^2)

So if I plug in the point coordinates (-2,3) to calculate the slope I get

(12 - (-24))/(-81+8) which is -36/73 so I know I am wrong b/c that just can't be the slope.

Where am I going wrong????

And jumping ahead ....the equation of the tangent will be
y-3 = (slope from above) (x- -2) and then solve for y if this is wrong let me know this too - thanks

this confused calculus beginner

2. Originally Posted by calcbeg
Okay I know I am doing something wrong but I can't see my mistake - please help me

"Find the equation of the tangent line to the curve
3x^2 - 3y^3+2yx^2=-45 at point (-2,3)

So first my goal is to find the derivative fox x so here I go....

6x - 9y^2 (dy/dx) + 4xy + 2x^2(dy/dx) = 0 **Forgot a dy in middle term

dy/dx (-9y^2 + 2x^2) = -6x -4xy

dy/dx = (-6x-4xy)/(-9y^2 + 2x^2)

So if I plug in the point coordinates (-2,3) to calculate the slope I get

(12 - (-24))/(-81+8) which is -36/73 so I know I am wrong b/c that just can't be the slope.

Where am I going wrong????

And jumping ahead ....the equation of the tangent will be
y-3 = (slope from above) (x- -2) and then solve for y if this is wrong let me know this too - thanks

this confused calculus beginner
You forgot your \frac{dy}{dx} on your middle term (*in red) See if that helps

3. I thought that b/c I wasn't calculating the derivative of y at that moment in time that I shouldn't put a dy/dx but I guess that implies one should go in wherever there is an x or y derivative being calculated??

Still a little confused but I will rework it this way and see what I get.

Thanks