What is the derivative of x^3 - x^2 + x + 1 from first principles?

Sunday, August 31, 2014

What is the derivative of x^3 - x^2 + x + 1 from first principles?

The derivative of a function f(x) from first principles is
f'(x) = $\displaystyle\lim_{{{h}\to{0}}}\frac{{{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}}}{{h}}$

Here f(x) =
x^3 - x^2 + x + 1

f'(x) = $\displaystyle\lim_{{{h}\to{0}}}{\left[{{\left({x}+{h}\right)}}^{{3}}-\right.}$
(x + h)^2 + (x + h) + 1 - (x^3 - x^2 + x + 1)]/h

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

=> f'(x) =
$\displaystyle\lim_{{{h}\to{0}}}{\left[{3}{{x}}^{{2}}{h}+{3}{x}{{h}}^{{2}}+{{h}}^{{3}}-{{h}}^{{2}}-{2}{x}{h}+\right.}$
h]/h

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

=> 3x^2 - 2x +
1

The derivative of f(x) = x^3 - x^2 + x + 1
is 3x^2 - 2x + 1

Help Notes

Sunday, August 31, 2014