How do you use the limit definition of the derivative to find the derivative of $f (x) = x^{3} + 2 x$ $f(x)=x^3+2x$ ?

Question

1519 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-13T13:11:13

$\frac{d}{d x} (x^{3} + 2 x) = 3 x^{2} + 2$ $d/dx (x^3+2x) = 3x^2+2$

By definition:

$(df)/dx = lim_(h->0) (f(x+h)-f(x))/h$

so:

$d/dx (x^3+2x) = lim_(h->0) ((x+h)^3+2(x+h)-x^3-2x)/h$

Expand the cube of the binomial and simplify:

$d/dx (x^3+2x) = lim_(h->0) (color(blue)(x^3)+3x^2h+3xh^2+h^3 +color(red)(2x)+2h-color(blue)(x^3)-color(red)(2x))/h$

$d/dx (x^3+2x) = lim_(h->0) (3x^2h+3xh^2+h^3 +2h)/h$

$d/dx (x^3+2x) = lim_(h->0) (cancel(h)(3x^2+3xh+h^2 +2))/cancel(h)$

$d/dx (x^3+2x) = lim_(h->0) 3x^2+3xh+h^2 +2$

$d/dx (x^3+2x) = 3x^2+2$

How do you use the limit definition of the derivative to find the derivative of f(x)=x^3+2xf(x)=x3+2xf(x)=x^3+2x?