What is the slope of the curve at the point (2,1)?

[carlarae]

Homework Statement

At the given point, find the slope of the curve, the line that is tangent to the curve, or the line that is normal to the curve, as requested. x^6y^6=64, normal at (2,1)

Homework Equations

The Attempt at a Solution

64/y^6
or 6x6y=0, that's as far as I am getting, totally lost.

 

[Simon Bridge]

You know the slope of the tangent at a point is the deriverative of the curve at that point right? So you need to find the deriverave of:

<br /> x^6 y^6 = 64<br />
at (x,y)=(2,1) and (guessing - you check) y is a function of x.

you can solve the equation for y, then find y' or find the implicit deriverative:
example

\frac{dy}{dx}:x^2y^2=4<br /> \frac{d}{dx} \left ( x^2y^2=4 \right )<br /> y^2\frac{d}{dx}x^2 + x^2\frac{d}{dx}y^2 = 0<br /> 2xy^2 + 2yx^2\frac{dy}{dx} = 0<br /> \frac{dy}{dx} = \frac{-2xy^2}{2yx^2} = -xy