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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?

(Also, what is hunterriverpei.comJaX?)

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

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\$\$

It"s valuable to store in psychic that two of the greatest hunterriverpei.comematical tricks are multiplication by \$1\$ and addition by \$0\$. Over I just multiplied through a an especially useful version of the number \$1\$.

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\$\$

\$\$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".