equation of the tangent line to the curve

**dat1611** · August 8th 2010, 01:26 PM

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

**Archie Meade** · August 8th 2010, 01:46 PM

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

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

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

**dat1611** · August 8th 2010, 02:09 PM

i dont understand how to do the derivative

**Plato** · August 8th 2010, 02:23 PM

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?

**pickslides** · August 8th 2010, 02:31 PM

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.

**dat1611** · August 8th 2010, 02:44 PM

how do you take d/dx of y^3

**pickslides** · August 8th 2010, 03:29 PM

Originally Posted by dat1611

how do you take d/dx of y^3

is $3y^2\frac{dy}{dx}$

Follow this example, you will find it very helpful.

Implicit Differentiation

**dat1611** · August 8th 2010, 05:39 PM

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

**skeeter** · August 8th 2010, 06:02 PM

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

**Archie Meade** · August 9th 2010, 03:32 AM

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.

**dat1611** · August 9th 2010, 02:01 PM

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)

**Archie Meade** · August 9th 2010, 02:07 PM

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.

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

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

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