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
Related Maths IB answers
All answers ▸Differentiate, from first principles, y=x^2
Find cos4x in terms of cosx.
A team of four is chosen from six married couples. If a husband and wife cannot both be on the team, in how many ways can the team be formed?
How to find a modulus and argument of w that is a quotient of z1 and z2 such that z1 = 1 + root(3)i and z2 = 1+ i using modulus-argument form?