Question

In Leibniz notation, we have:
dy/dx = (dy/du)(du/dx)

Lets say we have the function
f(x) = (x^2 + 1)^1/2 <-- sqrt function

What is the du/dx part?

Answer 1

2x

Answer 2

The du/dx part tell you to differentiate the argument of your function.
let u=sqrt(x) so you take the derivative of sqrt(x) and multiple it by the derivative of x.
dy/dx of sqrt(x)=1/(2sqrtx)*du/dx(x^2+1)--->(1/(2sqrt(x^2+1))*(2x)

Answer 3

Let u = x^2 + 1
Then f(u) = u^(1/2)
\[\frac{d}{dx}f(x) = \frac{d}{du}f(u)\frac{d}{dx}u = \frac{d}{du}u^{1/2}\frac{d}{dx}(x^{2}+1)\]First part is\[\frac{d}{du}u^{1/2} = \frac{1}{2}u^{-1/2}\]Second part is\[\frac{d}{dx}(x^2+1)=2x\]Now substitute back in for u, find the product, and simplify.