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QRAwarrior

  • 4 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?

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  1. mcmadlala
    • 4 years ago
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    2x

  2. giggles123
    • 4 years ago
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    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
    • 4 years ago
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    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.

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