Given: y + xy’= 5
Q: What is concavity of y = f(x) at (2, 6) ?
A couple ways to proceed here:
a. Solve for y’ first (in terms of x and y), explicitly differentiate y’ to obtain y’’, substituting ‘y as go, as it shows up.
b. Differentiate implicitly right away, to see if that speeds things up.
a. y + xy’= 5 ==> y’ = (5 - y)/x # Now, find y’’ using the Quotient Rule
y’’ = [x(-y’) - (5 - y)] / x² = [-xy’ - (5 - y)] / x² = [-x((5 - y)/x) - (5 - y)] / x² # Substitute y’ (above) into numerator
y’' = [(-2)(5 - y)] / x² … Observe that this is also (-2)(y’/x)
b. y + xy’= 5 ==> y’ = (5 - y)/x # (as above)… now Differentiate directly… Product Rule
y’ + xy’’ + y’ = 0 ==> y’’ = -2y’ / x = (-2)(5 - y)/x / x = (-2)(5 - y) / x² # Notice how much more quickly we got y''
Observations: We get the same answer for y’’, but “immediate” Implicit Differentiation followed by substitution “later" seems faster than Explicit Differentiation with early substitution.
Anyway, either way, at (2, 6), we get y’’ = (-2)(5 - y) / x² = (-2)(5 - 6) / 2² = 2/4 = 1/2 > 0
Ans. At (2, 6), y’’ > 0, so y is “Concave Up"
Fwiw, just a check:
By solving the Dif. Eq (by eyeballing), I find y = 2/x + 5 ==> y + xy’ = 2/x + 5 + x(-2/x²) = 2/x - 2/x + 5 = 5 ==> y’ = -2/x² ==> y’’ = 4/x³… At (2, 6), y’’ = 4/2³ = 1/2 > 0, and we now know that f(x) is Concave Up not only at x=2, but for all x > 0… Not needed for this exercise, but nice to confirm.
