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.


7 Answers 7


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.

  • 13
    $\begingroup$ 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. $\endgroup$
    – user22805
    Jun 29, 2012 at 22:47

$\sqrt x=x^{1/2}$, so you just use the power rule: the derivative is $\frac12x^{-1/2}$.

  • $\begingroup$ what about something like $2\sqrt{x}$ $\endgroup$
    – soniccool
    Jun 29, 2012 at 21:51
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    $\begingroup$ @soniccool: The derivative of $2f(x)$ is always twice the derivative of $f(x)$. So the derivative of $2\sqrt x$ is $2\cdot\frac12x^{-1/2}$. $\endgroup$
    – MJD
    Jun 29, 2012 at 21:52
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    $\begingroup$ @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 $$2\left(x^{1/2}\right)'=2\cdot\frac12x^{-1/2}=x^{-1/2}=\frac1{\sqrt x}\;.$$ $\endgroup$ Jun 29, 2012 at 22:19
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    $\begingroup$ Yes. Let $$y = 4\sqrt{x}$$ $$y' = 4(\frac{d}{dx} \sqrt{x})$$ $$y' = 4\cdot \frac{1}{2\sqrt{x}}$$ $$y' = \frac{2}{\sqrt{x}} $$ $\endgroup$
    – Joe
    Jun 29, 2012 at 22:49
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    $\begingroup$ @soniccool: You’re welcome. $\endgroup$ Jun 29, 2012 at 23:04

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: $$dx=d\sqrt x * \sqrt x + d\sqrt x * \sqrt x + d\sqrt x^2.$$

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:

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

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

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

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

Reference: Essence of Calculus, Chapter 3

  • $\begingroup$ 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. $\endgroup$
    – Nishant
    Oct 24, 2018 at 6:37
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    $\begingroup$ In essence, this approach can serve as a basis to prove the product rule in general $\endgroup$
    – imranfat
    Nov 11, 2018 at 17:58
  • $\begingroup$ 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)] $\endgroup$
    – Kymair Wu
    Aug 3, 2021 at 8:48

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 $$ \begin{align} y&=\sqrt{x}\\ y^2&=x\\ 2y\frac{\mathrm{d}y}{\mathrm{d}x}&=1\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=\frac{1}{2y}\\ &=\frac{1}{2\sqrt{x}}\tag{2} \end{align} $$


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,

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

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


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


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

  • $\begingroup$ Welcome to MSE. Did you read this before posting it? $\endgroup$ Oct 22, 2017 at 11:04
  • $\begingroup$ Oh I'm really sorry i don't know when posted it but, here is my answer \frac{1}{2\sqrt{x}$$ $\endgroup$
    – user488395
    Oct 22, 2017 at 11:11

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