You are watching: Derivative of the square root of x
I"m unsure just how to uncover the derivative the these and include them specifically 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 reality that if $f(x) = x^t$, climate $f"(x) = tx^t-1$.
Taking $t=1/2$, gives us the $f"(x) = dfrac12 x^-1/2$, i m sorry is the same as we obtained above.
Also, recall the $dfracd (c f(x))dx = c dfracdf(x)dx$. Hence, you can pull out the consistent and then identify it.
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answer Jun 29 "12 in ~ 21:52
user17762user17762
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$egingroup$ This is the finest answer here, since it doesn't assume that the power rule (which is basic to prove when the exponent is a positive integer) automatically uses when the exponent is not a positive integer. $endgroup$
–user22805
Jun 29 "12 at 22:47
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$sqrt x=x^1/2$, therefore you simply use the power 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 uncover the derivative the $f(x)=sqrt x$ is to use geometry. Imagine a square through side length $sqrt x$. Climate the area the the square is $x$. Now, let"s extend the square top top both sides by a tiny amount, $dsqrt x$. The brand-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 the the square (the orange locations in the photo above). The critical 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 2nd step come the last, flip the fountain on both sides of the equation.)
Reference: significance of Calculus, thing 3
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reply Dec 22 "17 in ~ 19:45
Armin MeisterhirnArmin Meisterhirn
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The Power preeminence says the $fracinvernessgangshow.netrmdinvernessgangshow.netrmdxx^alpha=alpha x^alpha-1$. Using this come $sqrtx=x^frac12$ gives$$eginalignfracinvernessgangshow.netrmdinvernessgangshow.netrmdxsqrtx&=fracinvernessgangshow.netrmdinvernessgangshow.netrmdxx^frac12\&=frac12x^-frac12\&=frac12sqrtx ag1endalign$$However, if you room uncomfortable applying the Power preeminence to a fractional power, consider applying implicit differentiation to$$eginaligny&=sqrtx\y^2&=x\2yfracinvernessgangshow.netrmdyinvernessgangshow.netrmdx&=1\fracinvernessgangshow.netrmdyinvernessgangshow.netrmdx&=frac12y\&=frac12sqrtx ag2endalign$$
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answer Jun 29 "12 in ~ 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 write-up the details implicit differentiation problem, it may help. The basic guideline of writing the square root together a fractional power and then making use of the power and also chain rule accordingly should be fine however. Also, remember that you can simply pull the end a constant when taking care of derivatives - view below.
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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
answer Jun 29 "12 in ~ 21:52
JoeJoe
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Use the product dominance $(fg)"=f"g+fg"$ and also take $f=sqrtx=g$. Then$$x"=1=(sqrtx cdot sqrtx)"=(sqrtx)"sqrtx+sqrtx(sqrtx)"=2sqrtx(sqrtx)".$$It complies with that $$(sqrtx)"=frac12sqrtx$$
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reply Nov 10 in ~ 9:22
Nicky HeksterNicky Hekster
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$sqrtx$Let $f(u)=u^1/2$ and $u=x$That"s $fracdfdu=frac12u^-1/2$ and also $fracdudx=1$But, by the chain preeminence $fracdydx=fracdfdu•fracdudx=frac12u^-1/2 •1=fracddxsqrtx$Finally$frac12sqrtx$
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edited Oct 22 "17 at 12:20
Daniel Fischer
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answered Oct 22 "17 in ~ 10:59
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