Use a factoring of a difference of cubes: a^3-b^3=(a-b)(a^2+ab+b^2), to help you calculate the derivative of f(x)=x^1/3 directly from the...

Tuesday, March 24, 2015

Use a factoring of a difference of cubes: a^3-b^3=(a-b)(a^2+ab+b^2), to help you calculate the derivative of f(x)=x^1/3 directly from the...

$\displaystyle{f{{\left({x}\right)}}}={{x}}^{{\frac{{1}}{{3}}}}$

$\displaystyle=\Rightarrow$
f'(x)= lim_(h->0) (f(x+h) - f(x))/ h

$\displaystyle=\Rightarrow$
f'(x)= lim_(h->0) ((x+h)^(1/3) - x^(1/3))/
h

$\displaystyle=\Rightarrow{\left({{a}}^{{3}}-{{b}}^{{3}}\right)}=$
(a-b)(a^2+ab+b^2)

$\displaystyle=\Rightarrow{\left({a}-{b}\right)}={\left({{a}}^{{\frac{{1}}{{3}}}}-{{b}}^{{\frac{{1}}{{3}}}}\right)}$
(a^(2/3) + (ab)^(1/3) + b^(2/3))

$\displaystyle=\Rightarrow{\left({{a}}^{{\frac{{1}}{{3}}}}-\right.}$
b^(1/3)) = (a-b)/ (a^(2/3) + (ab)^(1/3) +
b^(2/3))

$\displaystyle=\Rightarrow{{\left({x}+{h}\right)}}^{{\frac{{1}}{{3}}}}-{{x}}^{{\frac{{1}}{{3}}}}=\frac{{{x}+{h}-{x}}}{}$
((x+h)^(2/3) + ((x+h)x)^(1/3) + x^(2/3))

$\displaystyle=\Rightarrow$
lim_(h->0) f(x)= lim_(h->0) ((h/((x+h)^(2/3) + (x(x+h))^(1/3) +
x^(2/3)))/h)

$\displaystyle=\Rightarrow\lim_{{{h}\to{0}}}{f{{\left({x}\right)}}}=$
lim_(h->0) (1/((x+h)^(2/3) + (x(x+h))^(1/3) +
x^(2/3)))

$\displaystyle=\Rightarrow\lim_{{{x}\to{0}}}{f{{\left({x}\right)}}}=\frac{{1}}{{{{x}}^{{\frac{{2}}{{3}}}}}}$
+ x^(2/3) + x^(2/3)) = 1/(3x^(2/3))

$\displaystyle=\Rightarrow$
lim_(h-> 0) f(x) = 1/3 x^-(2/3)

$\displaystyle=\Rightarrow{f{'}}{\left({x}\right)}=$
1/3
x^-(2/3)

Help Notes

Tuesday, March 24, 2015