# Thread: On To Calculus - derivatives help

1. ## Re: On To Calculus - derivatives help

Originally Posted by rhickman
When I apply that rule I end up at

$(2)(\frac{1}{2})x^{0.5-1}$

Which gets me $x^{-0.5} or \frac{1}{\sqrt{x}}$
You are making a mistake

$f(x) = \sqrt{2x} = \sqrt{2} * \sqrt{x} = \sqrt{2} * x^{(1/2)} \implies$

$f'(x) = \sqrt{2} * \dfrac{1}{2} * x^{\{(1/2) - 1\}} = \dfrac{\sqrt{2}}{2} * x^{(-1/2)} = \dfrac{\sqrt{2}}{2} * \dfrac{1}{x^{(1/2)}} = \dfrac{\sqrt{2}}{2\sqrt{x}} = \dfrac{\sqrt{2}}{\sqrt{2} * \sqrt{2} * \sqrt{x}} = \dfrac{1}{\sqrt{2x}}.$

If you know the chain rule, it may be easier in terms of the algebra to do that.

$y = \sqrt{2x}.$

$u = 2x \implies \dfrac{du}{dx} = 2.$

$u = 2x \implies y = \sqrt{ u} = u^{1/2} \implies \dfrac{dy}{du} = \dfrac{1}{2} * u^{-1/2} = \dfrac{1}{2\sqrt{u}} = \dfrac{1}{2\sqrt{2x}}.$

$So\ \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} = 2 * \dfrac{1}{2\sqrt{2x}} = \dfrac{1}{\sqrt{2x}}.$

2. ## Re: On To Calculus - derivatives help

Ok I see my mistake now, thanks guys

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