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 ▸Let g (x) = 2x sin x . (a) Find g′(x) . (b) Find the gradient of the graph of g at x = π .
The quadratic equation 2x^2-8x+1 = 0 has roots a and b. Find the value of a + b and ab
Find the coordinates and determine the nature of the stationary points of curve y=(2/3)x^3+2x^2-6x+3
Prove 2^(n+2) + 3^(2n+1) is a multiple of 7 for all positive integers of n by mathematical induction.