Did you mean "with respect to x" ?
If so then I would square both sides
y^2 = xy + 9
2y dy/dx = x dy/dx + y
2y dy/dx - x dy/dx = y
dy/dx(2y-x) = y
dy/dx = y/(2y-x) such that √(xy-9) ≥ 0
if you want that expression to be in terms of x only, (like you typed)
we have a mess ahead
we would have to solve
y^2 - xy - 9 = 0 for y in terms of x
y = (x ± √(x^2 +36) )/2
and sub that into dy/dx to get only x's
so I will wait for your clarification before I start that messy substitution.
Find the derivative of the function y defined implicitly in terms of x.
y = sqrt(xy + 9)
Please help! Thanks! :)
1 answer
1 answer