$\displaystyle -3x^2+1xy-1y^3=-15$

I have to find the slope of the tangent line at the point of (-2,1)

Now from my work, the derivative of this function is

$\displaystyle dy/dx = (-6x+y)/(3+3y^2)$

Therefore the slope = 13/6

But the answer ain't right, why?

If anyone can help before I head off to the exam tomorrow morning at 9:30am I will really really really really appreicate it!!!

THANK YOU VERY MUCH!!!

$\displaystyle -3x^2+1xy-1y^3=-15$

I have to find the slope of the tangent line at the point of (-2,1)

Now from my work, the derivative of this function is

$\displaystyle dy/dx = (-6x+y)/(3+3y^2)$

Therefore the slope = 13/6

But the answer ain't right, why?

If anyone can help before I head off to the exam tomorrow morning at 9:30am I will really really really really appreicate it!!!

THANK YOU VERY MUCH!!!
Hello,

to derivate an implicite function use:
$\displaystyle \frac{dy}{dx}=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$

$\displaystyle \frac{\partial f}{\partial x}=-6x+y$
$\displaystyle \frac{\partial f}{\partial y}=-3y^2+x$

Greetings and good luck with your exam!

EB

3. earboth, he said implicit differentiation, not partial differentiation. hence it doesn't say
$\displaystyle f(x,y) = -3x^2 + xy - y^3 + 15$ or anything like that

anyway ttt, your answer shows you've got some idea how to go about things but its not quite right. we have
$\displaystyle -3x^2 + xy - y^3 = -15$
differentiating both sides... i will write $\displaystyle y'$ instead of $\displaystyle \frac{dy}{dx}$ to save time.
i will take you through term by term
for $\displaystyle -3x^2$, that differentiates to $\displaystyle -6x$, which i'm sure you already know.
now next i will look at $\displaystyle y^3$
$\displaystyle f(x) = y^3$
using the substitution $\displaystyle h=y$
$\displaystyle f(x) = h^3$
then
$\displaystyle \frac{df}{dh} = 3h^2$
also from the $\displaystyle h=y$ we have
$\displaystyle \frac{dh}{dx} = \frac{dy}{dx}$
so $\displaystyle \frac{df}{dh} \times \frac{dh}{dx} = 3y^2 \times \frac{dy}{dx}$
notice the dh's cancel
so differentiating $\displaystyle y^3$ implicitly with respect to x gives $\displaystyle 3y^2 y'$
which you might have also known.
where to do it quickly you are basically differentiating it with respect to y to get $\displaystyle 3y^2$. then multiplying it by $\displaystyle \frac{dy}{dx}$ to give $\displaystyle 3y^2y'$

now for the xy part you have to use the product rule
let
$\displaystyle g = x$
and
$\displaystyle h=y$
so
$\displaystyle g' = 1$
$\displaystyle h' = y'$
then $\displaystyle \frac{d(xy)}{dx} = gh' + hg' = xy' +y$
so to do it quickly when doing something with an x term times a y term you first differentiate with respect to x then multiply it by the y term... so to give you $\displaystyle 1 \times y$ then you add that to differentiating with respect to y implicitly multiplying by the x term which gives $\displaystyle x \times \frac{dy}{dx}$ or $\displaystyle xy'$
so in the original problem:
$\displaystyle -3x^2 + xy - y^3 = -15$
this becomes
$\displaystyle -6x + y + xy' - 3y^2y' = 0$
(the 15 differentiates to become 0 which you know)
so rearranging this gives
$\displaystyle xy' - 3x^2y' = 6x - y$
then
$\displaystyle y' = \frac{6x-y}{x-3y^2}$
then you can put in x=-2, y=1
$\displaystyle y' = 13/5$

by the way i apologize with my poor attempts at explaining whats going on in these examples. i just wanted to try and explain rather than merely give answers. i would appreciate it if somebody who knows things better would come along and clear them up for me.

4. Thank you, that is really helpful, I think I'm ready for the exam now.

I"m heading out for it, let's hope I will come back smiling.

Again, thanks for all of your helps!

*Update*

I just got back from school, the exam is much easier than the homework, I'm pretty confidence I made an A on that.

Thank you!