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Given $y = sqrt x$ and also nothing more, using the formula the a limit $$f"(x) = lim_h o0 fracf(x+h)-f(x)h$$ (that is, f element of x amounts to the border of h approaching zero through the equation ((f the the sum of x and h) minus (function that x)) end h)

how execute we convert (not evaluate) it right into Leibniz"s notation, $fracdydx$?

I"m having a lot of difficulties here, and also I hate taking radicals the end of denominators. Would certainly someone to walk me with this?

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Question 22 on page 107 the the publication from this difficulty here


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Here"s exactly how to advice the limit, in instance that"s what you"re asking:$$eginalignlim_h o 0 dfrac f(x+h)-f(x)h &= lim_h o 0 dfracsqrtx+h-sqrtxh \ &= lim_h o 0 dfracsqrtx+h-sqrtxhdfracsqrtx+h+sqrtxsqrtx+h+sqrtx \ &= lim_h o 0 dfrachh(sqrtx+h+sqrtx) \ &= lim_h o 0 dfrac1(sqrtx+h+sqrtx) \ &= dfrac12sqrtxendalign$$

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We desire to evaluate

$$lim_h ightarrow 0 fracsqrtx+h - sqrtxh$$

Multiply the top and bottom by $sqrtx+h + sqrtx$ to discover that this is same to

$$lim_h ightarrow 0 fracx+h-xh(sqrtx+h + sqrtx) = lim_h ightarrow 0 frachh(sqrtx+h + sqrtx)$$

Cancelling the $h$ native top and bottom, this is

$$lim_h ightarrow 0 frac1sqrtx+h+sqrtx = frac12sqrtx$$


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$$f"(x)=fracdydx$$ for $$y=f(x)$$ They room two different ways to write the exact same thing. No switch necessary, any much more than one would need to transform "$5$" come Roman numeral "V".

Just choose your weapon.




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Why don"t we find the general type of the power rule and then solve your problem? allow $$f(x)=x^n$$where $n$ is any real number, let"s find $fracdydx$ or similarly $f"$ utilizing The Equation:$$f"(x)=fracdydx= lim limits_h o 0 left( fracf(x+h)-f(x)h ight)$$$$=lim limits_h o 0 left( frac(x+h)^n - x^nh ight)$$$$=lim limits_h o 0 left( fracsum_r=0^n left( nchooserx^n-rh^r ight) - x^n h ight)$$$$=lim limits_h o 0 left( frac(x^n+nx^n-1h...+h^n) - x^nh ight)$$$$=lim limits_h o 0 left( fracnx^n-1h...+h^nh ight)$$$$=lim limits_h o 0 left( nx^n-1 ight) = nx^n-1$$Now allow $f(x)=sqrtx=x^frac12$$$f"(x)=fracdydx=frac12x^frac12-1=frac12x^-frac12=frac12x^frac12=frac12sqrtx$$