You are watching: How to take derivative of a square root

I"m unsure how to discover the derivative the these and include them particularly in something prefer implicit.

Let $f(x) = sqrtx$, then $$f"(x) = lim_h o 0 dfracsqrtx+h - sqrtxh = lim_h o 0 dfracsqrtx+h - sqrtxh imes dfracsqrtx+h + sqrtxsqrtx+h + sqrtx = lim_x o 0 dfracx+h-xh (sqrtx+h + sqrtx)\ = lim_h o 0 dfrachh (sqrtx+h + sqrtx) = lim_h o 0 dfrac1(sqrtx+h + sqrtx) = dfrac12sqrtx$$In general, you have the right to use the fact that if $f(x) = x^t$, climate $f"(x) = tx^t-1$.

Taking $t=1/2$, offers us the $f"(x) = dfrac12 x^-1/2$, i beg your pardon is the very same as we acquired above.

Also, recall that $dfracd (c f(x))dx = c dfracdf(x)dx$. Hence, you deserve to pull the end the constant and then differentiate it.

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answered Jun 29 "12 in ~ 21:52

user17762user17762

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$egingroup$ This is the ideal answer here, due to the fact that it doesn't assume that the power preeminence (which is straightforward to prove when the exponent is a positive integer) automatically applies when the exponent is no a optimistic integer. $endgroup$

–user22805

Jun 29 "12 at 22:47

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

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reply Jun 29 "12 in ~ 21:50

Brian M. ScottBrian M. Scott

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Another opportunity to find the derivative that $f(x)=sqrt x$ is to usage geometry. Imagine a square with side length $sqrt x$. Climate the area the the square is $x$. Now, let"s extend the square on both political parties by a tiny amount, $dsqrt x$. The new area included to the square is:$$dx=dsqrt x * sqrt x + dsqrt x * sqrt x + dsqrt x^2.$$

This is the amount of the sub-areas added on each side of the square (the orange locations in the photo above). The last term in the equation above is very tiny and deserve to be neglected. Thus:

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

$$fracdxdsqrt x=2 * sqrt x$$

$$fracdsqrt xdx=frac12*sqrt x$$

(To go from the 2nd step come the last, *flip* the fountain on both political parties of the equation.)

**Reference:** significance of Calculus, chapter 3

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answer Dec 22 "17 at 19:45

Armin MeisterhirnArmin Meisterhirn

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The Power dominion says that $fraccentregalilee.comrmdcentregalilee.comrmdxx^alpha=alpha x^alpha-1$. Applying this come $sqrtx=x^frac12$ gives$$eginalignfraccentregalilee.comrmdcentregalilee.comrmdxsqrtx&=fraccentregalilee.comrmdcentregalilee.comrmdxx^frac12\&=frac12x^-frac12\&=frac12sqrtx ag1endalign$$However, if you room uncomfortable applying the Power ascendancy to a spring power, consider applying implicit differentiation to$$eginaligny&=sqrtx\y^2&=x\2yfraccentregalilee.comrmdycentregalilee.comrmdx&=1\fraccentregalilee.comrmdycentregalilee.comrmdx&=frac12y\&=frac12sqrtx ag2endalign$$

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answered Jun 29 "12 at 22:04

robjohn♦robjohn

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Let $f(x) = sqrtx = x^1/2$.

$$f"(x) = frac12 x ^-1/2$$

$$f"(x) = frac12x^1/2 = frac12sqrtx$$

If you article the specific implicit differentiation problem, it might help. The basic guideline of writing the square root as a fountain power and also then making use of the power and chain rule appropriately should be fine however. Also, remember the you have the right to simply pull out a continuous when taking care of derivatives - view below.

See more: What Does Lulu Mean In Hawaiian Word Is, What Does Lulu Mean In Swahili

If $g(x) = 2sqrtx = 2x^1/2$. Then,

$$g"(x) = 2cdotfrac12x^-1/2$$

$$g"(x) = frac1x^1/2 = frac1sqrtx$$

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edited Jun 29 "12 at 22:08

answered Jun 29 "12 at 21:52

JoeJoe

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$sqrtx$Let $f(u)=u^1/2$ and also $u=x$That"s $fracdfdu=frac12u^-1/2$ and $fracdudx=1$But, by the chain preeminence $fracdydx=fracdfdu•fracdudx=frac12u^-1/2 •1=fracddxsqrtx$Finally$frac12sqrtx$

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edited Oct 22 "17 in ~ 12:20

Daniel Fischer

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answer Oct 22 "17 in ~ 10:59

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