What would be the derivative the square roots? For example if I have $2 sqrtx$ or $sqrtx$.

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I"m unsure exactly how to discover the derivative the these and also include them especially in something choose implicit.


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

Taking $t=1/2$, gives us the $f"(x) = dfrac12 x^-1/2$, which is the same as we acquired above.

Also, recall the $dfracd (c f(x))dx = c dfracdf(x)dx$. Hence, you can pull out the constant and then differentiate it.


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reply Jun 29 "12 in ~ 21:52
user17762user17762
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$egingroup$ This is the ideal answer here, since it doesn't assume that the power dominion (which is straightforward to prove as soon as the exponent is a optimistic integer) automatically uses when the exponent is no a optimistic integer. $endgroup$
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Jun 29 "12 at 22:47
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$sqrt x=x^1/2$, therefore you simply use the strength rule: the derivative is $frac12x^-1/2$.


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answer Jun 29 "12 at 21:50
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Brian M. ScottBrian M. Scott
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Another opportunity to find the derivative the $f(x)=sqrt x$ is to use geometry. Imagine a square v side size $sqrt x$. Then the area that the square is $x$. Now, let"s expand the square ~ above both political parties by a little amount, $dsqrt x$. The brand-new area added to the square is:$$dx=dsqrt x * sqrt x + dsqrt x * sqrt x + dsqrt x^2.$$

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This is the amount of the sub-areas included on every side the the square (the orange areas in the picture above). The last term in the equation over is very small 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 second step come the last, flip the fountain on both sides of the equation.)

Reference: significance of Calculus, chapter 3


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answer Dec 22 "17 in ~ 19:45
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Armin MeisterhirnArmin Meisterhirn
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The Power dominance says the $fracchrischona2015.orgrmdchrischona2015.orgrmdxx^alpha=alpha x^alpha-1$. Applying this come $sqrtx=x^frac12$ gives$$eginalignfracchrischona2015.orgrmdchrischona2015.orgrmdxsqrtx&=fracchrischona2015.orgrmdchrischona2015.orgrmdxx^frac12\&=frac12x^-frac12\&=frac12sqrtx ag1endalign$$However, if you space uncomfortable applying the Power dominance to a fountain power, consider using implicit differentiation to$$eginaligny&=sqrtx\y^2&=x\2yfracchrischona2015.orgrmdychrischona2015.orgrmdx&=1\fracchrischona2015.orgrmdychrischona2015.orgrmdx&=frac12y\&=frac12sqrtx ag2endalign$$


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answer Jun 29 "12 in ~ 22:04
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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 particular implicit differentiation problem, it might help. The basic guideline of creating the square root together a fountain power and then making use of the power and also chain rule as necessary should it is in fine however. Also, remember the you deserve to simply pull the end a continuous when dealing with derivatives - see below.

See more: I Hope Everything Is Going Well With You, I Hope Everything Is Going Well For You

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 in ~ 22:08
reply Jun 29 "12 in ~ 21:52
JoeJoe
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$sqrtx$Let $f(u)=u^1/2$ and $u=x$That"s $fracdfdu=frac12u^-1/2$ and $fracdudx=1$But, through the chain preeminence $fracdydx=fracdfdu•fracdudx=frac12u^-1/2 •1=fracddxsqrtx$Finally$frac12sqrtx$


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edited Oct 22 "17 at 12:20
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