# Thread: equation of the tangent line to the curve

1. ## equation of the tangent line to the curve

x^3+3xy+y^3=1 at the point (1,0)

i solved to get y =sqrt(x^3+3x-1)

then i think you take the derivative of that and set it equal to 0

im not sure how i procede

2. Originally Posted by dat1611
x^3+3xy+y^3=1 at the point (1,0)

i solved to get y =sqrt(x^3+3x-1)

this would be the case if the function was x^3+3x-y^2=1

then i think you take the derivative of that and set it equal to 0

you would do that to find a local maximum or minimum.

im not sure how i procede
You need "implicit" differentiation.

$x^3+3xy+y^3=1$

Differentiate both sides, realising that the derivative of a constant is zero.

$3x^2+3\frac{d}{dx}(xy)+\frac{d}{dx}y^3=0$

xy is a product and you need the chain rule for the "y-cubed" term.

Then, when you've got that, use x=1 and y=0.
The derivative at that point is the slope of the tangent.

Then the equation of the tangent is

$y-0=m(x-1)$

$y=\frac{dy}{dx}(x-1)$

3. i dont understand how to do the derivative

4. Originally Posted by dat1611
i dont understand how to do the derivative
If you do not understand the derivative, then why are you trying to do a problem that involves derivatives?

5. Originally Posted by dat1611
x^3+3xy+y^3=1 at the point (1,0)

i solved to get y =sqrt(x^3+3x-1)

then i think you take the derivative of that and set it equal to 0

im not sure how i procede
You should reivew the chain rule to get $\frac{dy}{dx}$ here.

6. how do you take d/dx of y^3

7. Originally Posted by dat1611
how do you take d/dx of y^3
is $3y^2\frac{dy}{dx}$

Implicit Differentiation

8. i got -x^2/y^2 but if i plug in x and y it undefined

9. Originally Posted by dat1611
i got -x^2/y^2 but if i plug in x and y it undefined
your derivative is incorrect. it should be ...

$\frac{dy}{dx} = -\frac{x^2+y}{x+y^2}$

you need to review the method for taking implicit derivatives ...

Implicit Differentiation

10. Originally Posted by dat1611
i got -x^2/y^2 but if i plug in x and y it undefined
You forgot to differentiate the "product term" 3xy using the product rule of differentiation.

11. so i plug in the values for x any y and got slope being negative 1 so is the answer just y-0=-1(x-1)

12. Originally Posted by dat1611
so i plug in the values for x any y and got slope being negative 1 so is the answer just y-0=-1(x-1)
Yep,

here's a way to "think of" the chain rule.

$\displaystyle\huge\frac{d}{dy}y^3=3y^2$

$\displaystyle\huge\frac{d}{dx}y^3=\frac{dy}{dy}\fr ac{d}{dx}y^3=\frac{dy}{dx}\frac{d}{dy}y^3$