Calculus
posted by Kelly .
Both x and y denote functions of t that are related by the given equation. Use this equation and the given derivative information to find the specified derivative.
Equation:35x^2+20y^2=1
Given that dx/dt=4, find dy/dt
when (x,y)=(1/7sqrt2, 1/5sqrt2)

70 x dx/dt + 40 y dy/dt = 0
280 x + 40 y dy/dy = 0
280/7sqrt2 + 40/5sqrt2 dy/dt = 0