## QRAwarrior 3 years ago 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?

2x

2. giggles123

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)

3. dmancine

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.