Derivatives

May 2010
8
0
Find the fluxion according the definition

\(\displaystyle Ln^2(2+3x)\)

\(\displaystyle \sqrt[3]{(1+4x)}\)

\(\displaystyle x^4 +3x +1\)

Please post with solving not only answers.
 
Last edited by a moderator:
May 2010
8
0
need to find derative according definition
 
Jun 2009
806
275
Find the fluxion according the definition

\(\displaystyle Ln^2(2+3x)\)

\(\displaystyle \sqrt[3]{(1+4x)}\)

\(\displaystyle x^4 +3x +1\)

Please post with solving not only answers.
f(x) = Ln^2(2+3x)

f(x+h) = Ln^2{2+3(x+h)}

f(x+h) - f(x) =Ln^2{2+3(x+h)} - Ln^2(2+3x)

\lim_{h\to\0}\frav{f(x+h) - f(x)}{h}= \lim_{h\to\0}\frac{Ln^2{2+3(x+h)} - Ln^2(2+3x)}{h}

\lim_{h\to\0}\frav{f(x+h) - f(x)}{h} =\lim_{h\to\0}\frac{Ln[{2+3(x+h)} + Ln(2+3x)][Ln{2+3(x+h)} - Ln(2+3x)}{h}

\lim_{h\to\0}\frav{f(x+h) - f(x)}{h} =\lim_{h\to\0}\frac{Ln[{2+3(x+h)} + Ln(2+3x)][Ln(1+\frac{h}{2+3x})]}{h}

\lim_{h\to\0}\frav{f(x+h) - f(x)}{h} =\lim_{h\to\0}[Ln[{2+3(x+h)} + Ln(2+3x)][Ln(1+\frac{3h}{2+3x})^\frac{2+3x}{3h}][{3/(2+3x)}

Now find the limits to get the result.

Similarly proceed with other two.
 
Last edited:
May 2010
8
0
\lim_{h\to\0}\frav{f(x+h) - f(x)}{h} =\lim_{h\to\0}\frac{Ln[{2+3(x+h)} + Ln(2+3x)][Ln(1+\frac{h}{2+3x})]}{h}

Can you explain how do you get

[Ln(1+\frac{h}{2+3x})]
 
May 2010
8
0
And how to count the limits to get answer. ;)
 
Jun 2009
806
275
And how to count the limits to get answer. ;)
[Ln{2+3(x+h)} - Ln(2+3x)}{h}

[ln(2 + 3x + 3h) - ln(2 + 3x)]/h

= [ln( 1 + 3h/(2+3x)]/h

Lim as x tends to zero (1+x)^1/x = e.

let t = 3h/(2+3x)
1/t*ln(1+t) = ln(1+t)^1/t

[ln( 1 + 3h/(2+3x)]/h ={ ln[( 1 + 3h/(2+3x)]^(2+3x)/3h}*3h/(2+3x)h

Limit of this is 3/(2+3x)
 
Jun 2009
806
275
But the derative is 6LN(2+3x)/(2+3x)
\lim_{h\to\0}\frac{Ln[{2+3(x+h)} + Ln(2+3x)][Ln{2+3(x+h)} - Ln(2+3x)}{h}


lim_{h\to\0}\frac{Ln[{2+3(x+h)} + Ln(2+3x)]

If you h = 0 in the above expression you get

2ln(2+3x)
 
May 2010
8
0
How to solve second problem? I know steps, but I dont know what to do with the root