Differentiating the Chain Rule to get Second Derivative?

Question

If y is given by a composite function of x, use the Chain rule to show that

d^2y/dx^2 = d^2y/du^2 * (du/dx)^2 + dy/dy * d^2u/dx^2

How does one differentiate dy/dx = dy/du * du/dx to get the above?

Answer 1

d(dy/du)/dx = d(dy/du)/du * du/dx = d^2 y/du^2 * du/dx

d^2 y / dx^2 = d(dy/dx)/dx = d(dy/du * du/dx)/dx =
= d(dy/du)/dx * du/dx + d(du/dx)/dx * dy/du =
= d^2 y/du^2 * du/dx * du/dx + d^2 u / dx^2 * dy/du =
= d^2y/du^2 * (du/dx)^2 + dy/du * d^2u/dx^2

Answer 2

Start with the first derivative:

dy/dx = dy/du du/dx

Now take another derivative

d^2y/dx^2 = d(dy/du du/dx)/dx = d(dy/du)/dx du/dx + dy/du d^2u/dx^2

dy^2/dx^2 = d(dy/dx)/du du/dx + d^2u/dx^2 dy/du because order of differentiation doesn't matter

d^2y/dx^2 = d(dy/du du/dx)/du du/dx +d^2u/dx^2 dy/du

= d^2y/du^2 (du/dx )^2 +du^2/dx^2 dy/du

Answer 3

Remember the product rule.... The derivative of the first times the second + the derivative of the second times the first (and then the chain rule). So, x'e^5x + x(e^5x)' e^5x + x(e^5x times (5x)')....there's the chain rule e^5x + x(e^5x times 5) e^5x + 5xe^5x Hope this helps!