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What would be the derivative of square roots? For example if I have $2 \sqrt{x}$ or $\sqrt{x}$.

I'm unsure how to find the derivative of these and include them especially in something like implicit.

Irregular User
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soniccool
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7 Answers7

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Let $f(x) = \sqrt{x}$, then $$f'(x) = \lim_{h \to 0} \dfrac{\sqrt{x+h} - \sqrt{x}}{h} = \lim_{h \to 0} \dfrac{\sqrt{x+h} - \sqrt{x}}{h} \times \dfrac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} = \lim_{x \to 0} \dfrac{x+h-x}{h (\sqrt{x+h} + \sqrt{x})}\\ = \lim_{h \to 0} \dfrac{h}{h (\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \dfrac1{(\sqrt{x+h} + \sqrt{x})} = \dfrac1{2\sqrt{x}}$$ In general, you can use the fact that if $f(x) = x^{t}$, then $f'(x) = tx^{t-1}$.

Taking $t=1/2$, gives us that $f'(x) = \dfrac12 x^{-1/2}$, which is the same as we obtained above.

Also, recall that $\dfrac{d (c f(x))}{dx} = c \dfrac{df(x)}{dx}$. Hence, you can pull out the constant and then differentiate it.

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    This is the best answer here, because it doesn't assume that the power rule (which is easy to prove when the exponent is a positive integer) automatically applies when the exponent is NOT a positive integer. –  Jun 29 '12 at 22:47
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$\sqrt x=x^{1/2}$, so you just use the power rule: the derivative is $\frac12x^{-1/2}$.

Brian M. Scott
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  • what about something like $2\sqrt{x}$ – soniccool Jun 29 '12 at 21:51
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    @soniccool: The derivative of $2f(x)$ is always twice the derivative of $f(x)$. So the derivative of $2\sqrt x$ is $2\cdot\frac12x^{-1/2}$. – MJD Jun 29 '12 at 21:52
  • Ah alright i got it – soniccool Jun 29 '12 at 21:52
  • So why does $2\sqrt{x}$ = $\frac{1}{\sqrt{x}}$ – soniccool Jun 29 '12 at 22:13
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    @soniccool: You mean $\left(2\sqrt x\right)'=\frac1{\sqrt x}$. For the same reason that the derivative of $2x^3$ is $6x^2$: the derivative of $af(x)$ is $af'(x)$. In this case you have the derivative of $2x^{1/2}$, which is $$2\left(x^{1/2}\right)'=2\cdot\frac12x^{-1/2}=x^{-1/2}=\frac1{\sqrt x};.$$ – Brian M. Scott Jun 29 '12 at 22:19
  • Okay i get it so basically you bring out the 2 and then 2 becomes 0 right? – soniccool Jun 29 '12 at 22:41
  • So would $4\sqrt{x}$ be the same thing? – soniccool Jun 29 '12 at 22:41
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    Yes.

    Let $$y = 4\sqrt{x}$$

    $$y' = 4(\frac{d}{dx} \sqrt{x})$$

    $$y' = 4\cdot \frac{1}{2\sqrt{x}}$$

    $$y' = \frac{2}{\sqrt{x}} $$

     – Joe Jun 29 '12 at 22:49
  • @soniccool: The $2$ does not become $0$. Do you mean that if you differentiate $2x^{1/2}$ as a product, you get $2\cdot\frac12x^{-1/2}+0\cdot x^{1/2}$? That’s true, and you can differentiate $2x^{1/2}$ as a product if you really want to, but it’s making unnecessary work for yourself. Don’t make things hard for yourself: use the fact that for any constant $a$ and any differentiable function $f(x)$, $\left(af(x)\right)'=af'(x)$. – Brian M. Scott Jun 29 '12 at 23:03
  • AH okay im an idiot i got it. – soniccool Jun 29 '12 at 23:03
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    @soniccool: You’re welcome. – Brian M. Scott Jun 29 '12 at 23:04
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Another possibility to find the derivative of $f(x)=\sqrt x$ is to use geometry. Imagine a square with side length $\sqrt x$. Then the area of the square is $x$. Now, let's extend the square on both sides by a small amount, $d\sqrt x$. The new area added to the square is: $$dx=d\sqrt x \times \sqrt x + d\sqrt x \times \sqrt x + d\sqrt x^2.$$

enter image description here

This is the sum of the sub-areas added on each side of the square (the orange areas in the picture above). The last term in the equation above is very small and can be neglected. Thus:

$$dx=2\times d\sqrt x \times \sqrt x$$

$$\frac{dx}{d\sqrt x}=2 \times \sqrt x$$

$$\frac{d\sqrt x}{dx}=\frac{1}{2\times\sqrt x}$$

(To go from the second step to the last, flip the fractions on both sides of the equation.)

Reference: Essence of Calculus, Chapter 3

IraeVid
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Armin Meisterhirn
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  • Nice! I came from the same source i.e 3Blue1Brown's Essense of Calculus series. But do we get any geometric interpretation as such as in the case of x^2 function? This feels more like an algebraic manipulation that leads to the same value. – Nishant Oct 24 '18 at 06:37
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    In essence, this approach can serve as a basis to prove the product rule in general – imranfat Nov 11 '18 at 17:58
  • I kinda feel bad about "The last term in the equation above is very small and can be neglected."

    Because we only have dx approaches to zero, instead of d(x^1/2)]

     – Kymair Wu Aug 03 '21 at 08:48
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The Power Rule says that $\frac{\mathrm{d}}{\mathrm{d}x}x^\alpha=\alpha x^{\alpha-1}$. Applying this to $\sqrt{x}=x^{\frac12}$ gives $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\sqrt{x} &=\frac{\mathrm{d}}{\mathrm{d}x}x^{\frac12}\\ &=\frac12x^{-\frac12}\\ &=\frac{1}{2\sqrt{x}}\tag{1} \end{align} $$ However, if you are uncomfortable applying the Power Rule to a fractional power, consider applying implicit differentiation to $$ \begin{align} y&=\sqrt{x}\\ y^2&=x\\ 2y\frac{\mathrm{d}y}{\mathrm{d}x}&=1\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=\frac{1}{2y}\\ &=\frac{1}{2\sqrt{x}}\tag{2} \end{align} $$

robjohn
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Let $f(x) = \sqrt{x} = x^{1/2}$.

$$f'(x) = \frac{1}{2} x ^{-1/2}$$

$$f'(x) = \frac{1}{2x^{1/2}} = \frac{1}{2\sqrt{x}}$$

If you post the specific implicit differentiation problem, it may help. The general guideline of writing the square root as a fractional power and then using the power and chain rule appropriately should be fine however. Also, remember that you can simply pull out a constant when dealing with derivatives - see below.

If $g(x) = 2\sqrt{x} = 2x^{1/2}$. Then,

$$g'(x) = 2\cdot\frac{1}{2}x^{-1/2}$$

$$g'(x) = \frac{1}{x^{1/2}} = \frac{1}{\sqrt{x}}$$

Joe
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Use the product rule $(fg)'=f'g+fg'$ and take $f=\sqrt{x}=g$. Then $$x'=1=(\sqrt{x} \cdot \sqrt{x})'=(\sqrt{x})'\sqrt{x}+\sqrt{x}(\sqrt{x})'=2\sqrt{x}(\sqrt{x})'.$$ It follows that $$(\sqrt{x})'=\frac{1}{2\sqrt{x}}$$

Nicky Hekster
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$\sqrt{x}$ Let $f(u)=u^{1/2}$ and $u=x$ That's $\frac{df}{du}=\frac{1}{2}u^{-1/2}$ and $\frac{du}{dx}=1$ But, by the chain rule $\frac{dy}{dx}=\frac{df}{du}•\frac{du}{dx} =\frac{1}{2}u^{-1/2} •1 =\frac{d}{dx}\sqrt{x}$ Finally $\frac{1}{2\sqrt{x}}$

Daniel Fischer
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user488395
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