Given 2x^2-3y^2=2, find the two values of dy/dx when x=5.
First solve for the exact point on the line by substituting 5 into the original equation. You should get y=+-4.
Now implicitly differentiate the equation: 4x-6y(dy/dx)=0. Rearranging this will yield the following: dy/dx=(2x)/(3y). Because we only have one value of x, let's substitute this into the derivative first: dy/dx=10/3y. Now we can individually substitute the two y values to get the two values of dy/dx. dy/dx = 10/12 = 5/6, dy/dx = -10/12 = -5/6 These are the two values of dy/dx when x=5.
KU
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