# Derivative Of 1 Over Square Root Of X ) = Square Root Of X, Finding The Derivative Of The Square Root Of X

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

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

Let $f(x) = sqrt{x}$, then $$f”(x) = lim_{h o 0} dfrac{sqrt{x+h} – sqrt{x}}{h} = lim_{h o 0} dfrac{sqrt{x+h} – sqrt{x}}{h} imes dfrac{sqrt{x+h} + sqrt{x}}{sqrt{x+h} + sqrt{x}} = lim_{x o 0} dfrac{x+h-x}{h (sqrt{x+h} + sqrt{x})}\ = lim_{h o 0} dfrac{h}{h (sqrt{x+h} + sqrt{x})} = lim_{h o 0} dfrac1{(sqrt{x+h} + sqrt{x})} = dfrac1{2sqrt{x}}$$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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answered Jun 29 “12 at 21:52

user17762user17762

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$egingroup$ 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. $endgroup$

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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, $dsqrt x$. The new area added to the square is:$$dx=dsqrt x * sqrt x + dsqrt x * sqrt x + dsqrt x^2.$$

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:

$$dx=2*dsqrt x * sqrt x$$

$$frac{dx}{dsqrt x}=2 * sqrt x$$

$$frac{dsqrt x}{dx}=frac{1}{2*sqrt x}$$

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

**Reference:** Essence of Calculus, Chapter 3

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answered Dec 22 “17 at 19:45

Armin MeisterhirnArmin Meisterhirn

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The Power Rule says that $frac{

adechworld.comrm{d}}{

adechworld.comrm{d}x}x^alpha=alpha x^{alpha-1}$. Applying this to $sqrt{x}=x^{frac12}$ gives$$egin{align}frac{

adechworld.comrm{d}}{

adechworld.comrm{d}x}sqrt{x}&=frac{

adechworld.comrm{d}}{

adechworld.comrm{d}x}x^{frac12}\&=frac12x^{-frac12}\&=frac{1}{2sqrt{x}} ag{1}end{align}$$However, if you are uncomfortable applying the Power Rule to a fractional power, consider applying implicit differentiation to$$egin{align}y&=sqrt{x}\y^2&=x\2yfrac{

adechworld.comrm{d}y}{

adechworld.comrm{d}x}&=1\frac{

adechworld.comrm{d}y}{

adechworld.comrm{d}x}&=frac{1}{2y}\&=frac{1}{2sqrt{x}} ag{2}end{align}$$

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answered Jun 29 “12 at 22:04

robjohn♦robjohn

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

$$f”(x) = frac{1}{2} x ^{-1/2}$$

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

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.

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

$$g”(x) = 2cdotfrac{1}{2}x^{-1/2}$$

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

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edited Jun 29 “12 at 22:08

answered Jun 29 “12 at 21:52

JoeJoe

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

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edited Oct 22 “17 at 12:20

Daniel Fischer

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answered Oct 22 “17 at 10:59

user488395user488395

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