Graph tangent equation

**Revy** · March 1st 2010, 05:55 PM

I have graph $x^4+y^4+2=4xy^3$ and I need to find tanget equation at $(1;1)$ (on a graph) point.

I can't seem to get what to do with $x$ s and $y$ s

can't convey neither of those

Do I need to do partial fluxion for $x$ and then for $y$ ? (well... these ones: $Fx'=...$ and $Fy'=...$ )

If yes, what should I do next with those 2 functions?

**Haven** · March 1st 2010, 10:30 PM

For implicit differentiation or Partial Fluxion as you call it. You apply the following:

1) Take the derivative with respect to x on both sides
$\frac{d}{dx}(x^4+y^4+2) = \frac{d}{dx}(4xy^3)$

2)Any where you see a term with no y's in it, differentiate like normal
$4x^3 + \frac{d}{dx}y^4 = \frac{d}{dx}(4xy^3)$

3) Now in the terms with y's in them, we differentiate with respect to y but keep the term \frac{dy}{dx}. This is essentially the chain rule.
$4x^3 + \frac{dy}{dx}4y^3 = \frac{d}{dx}(4xy^3)$

4) Now in this case we'll have to apply the product rule to the right hand side of the equation
$4x^3 + \frac{dy}{dx}4y^3 = 4y^3 + \frac{dy}{dx}12xy^2$

Now we solve for $\frac{dy}{dx}$

$\frac{dy}{dx}(4y^3 - 12xy^2) = 4y^3 - 4x^3$
$\frac{dy}{dx} = \frac{4y^3 - 4x^3}{4y^3 - 12xy^2}$

Now we substitute in the point (1,1) and we get:
$\frac{dy}{dx} = \frac{4-4}{4-12} = 0$
So the slope of the tangent line at that point is 0.

**Revy** · March 1st 2010, 11:52 PM

Tanget equation is just $y=1$ since it's horizontal for $x$ axis?

Thread: Graph tangent equation

LinkBack

Thread Tools

Display

Graph tangent equation

Similar Math Help Forum Discussions

Find the equation of the tangent line(s) to the graph

Equation of tangent line to the graph

equation for the line tangent to graph

Equation For The Line Tangent To The Graph

Find an equation for the line tangent to the graph