Calculus 3 problem partial derivatives?

Comments

• kb

  May 2021

  Let u = a^2 - b^2 and v = 4ab.

  By the Chain Rule,

  ∂c/∂a = ∂f/∂u ∂u/∂a + ∂f/∂v ∂v/∂a

  .........= 2a ∂f/∂u + 4b ∂f/∂v.

  ∂²c/∂a² = (∂/∂a)(∂c/∂a)

  ...........= (∂/∂a)[2a ∂f/∂u + 4b ∂f/∂v]

  ...........= [2 * ∂f/∂u + 2a * (∂/∂a)(∂f/∂u)] + 4b (∂/∂a)(∂f/∂v)

  ...........= 2 ∂f/∂u + 2a [2a ∂²f/∂u² + 4b ∂²f/∂v∂u] + 4b [2a ∂²f/∂u∂v + 4b ∂²f/∂v²]

  ...........= 2 ∂f/∂u + 4a² ∂²f/∂u² + 16ab ∂²f/∂u∂v + 16b² ∂²f/∂v².

  I hope this helps!
• anon-e-moose-

  May 2021

  c = f(a²-b², 4ab)

  let u(a,b) = a²-b²

  let v(a,b) = 4ab

  ∂u/∂a = 2a

  ∂v/∂a = 4b

  ∂c/∂a = ∂f(u,v)/∂a

  ∂c/∂a = ∂f(u,v)/∂u * ∂u/∂a + ∂f(u,v)/∂v * ∂v/∂a

  ∂c/∂a = ∂f(u,v)/∂u * 2a + ∂f(u,v)/∂v * 4b

  ∂²c/∂a² = ∂²f(u,v)/∂u∂a * 2a + ∂²f(u,v)/∂v∂a * 4b

  ∂²c/∂a² = [∂²f(u,v)/∂²u * ∂u/∂a + ∂²f(u,v)/∂u∂v * ∂v/∂a] * 2a + [∂²f(u,v)/∂v∂u * ∂u/∂a + ∂²f(u,v)/∂²v * ∂v/∂a] * 4b

  ∂²c/∂a² = [∂²f(u,v)/∂²u * 2a + ∂²f(u,v)/∂u∂v * 4b] * 2a + [∂²f(u,v)/∂v∂u * 2a + ∂²f(u,v)/∂²v * 4b] * 4b

  ∂²c/∂a² = ∂²f(u,v)/∂²u * 4a² + ∂²f(u,v)/∂u∂v * 8ab + ∂²f(u,v)/∂v∂u * 8ab + ∂²f(u,v)/∂²v * 16b²

  ∂²c/∂a² = ∂²f(u,v)/∂²u * 4a² + ∂²f(u,v)/∂u∂v * 16ab + ∂²f(u,v)/∂²v * 16b²

  ∂²c/∂a² = 4 * [∂²f(u,v)/∂²u * a² + ∂²f(u,v)/∂u∂v * 4ab + ∂²f(u,v)/∂²v * 4b²]

  ∂²c/∂a² = 4 * [∂/∂u * a + ∂/∂v * 2b]² f(u,v)