Answered: "What would be the derivative of square roots?..."

Colton Gregory

Open question

2022-08-20

What would be the derivative of square roots? For example if I have

2 \sqrt{x}

\sqrt{x}

Answer & Explanation

suprakarmay5

Beginner2022-08-21Added 8 answers

Let $f (x) = \sqrt{x}$ , then
$f^{'} (x) = lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h} = lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h} \times \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}}$
$= lim_{x \to 0} \frac{x + h - x}{h (\sqrt{x + h} + \sqrt{x})}$
$= lim_{h \to 0} \frac{h}{h (\sqrt{x + h} + \sqrt{x})}$
$= lim_{h \to 0} \frac{1}{(\sqrt{x + h} + \sqrt{x})}$
$= \frac{1}{2 \sqrt{x}}$
In general, you can use the fact that if $f (x) = x^{t} ask, then f^{'} (x) = t x^{t - 1}$
Taking $t = \frac{1}{2}$ , gives us that $f^{'} (x) = \frac{1}{2} x^{- \frac{1}{2}}$ , which is the same as we obtained above.
Also, recall that $\frac{d (c f (x))}{d x} = c \frac{d f (x)}{d x}$ . Hence, you can pull out the constant and then differentiate it.

musouorochidf

Beginner2022-08-22Added 2 answers

The Power Rule says that

\frac{d}{d x} x^{α} = α x^{α - 1}

. Applying this to

\sqrt{x} = x^{\frac{1}{2}}

gives

\frac{d}{d x} \sqrt{x} = \frac{d}{d x} x^{\frac{1}{2}}

= \frac{1}{2} x^{- \frac{1}{2}}

= \frac{1}{2 \sqrt{x}}

However, if you are uncomfortable applying the Power Rule to a fractional power, consider applying implicit differentiation to

y = \sqrt{x}

y^{2} = x

2 y \frac{d y}{d x} = 1

\frac{d y}{d x} = \frac{1}{2 y}

= \frac{1}{2 \sqrt{x}}

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