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.