Answer
$ f’(x) = 3x^2 $
The equation of the tangent line: $y = 0 $
Work Step by Step
$f(x) = x^3 $
Find the derivative:
$ f’(x) = \lim\limits_{ Δx\to 0} \frac{f(x+Δx) - f(x)}{Δx} $
$ f’(x) = \lim\limits_{ Δx\to 0} \frac{(x+Δx)^3 - x^3}{Δx} $
$ f’(x) = \lim\limits_{ Δx\to 0} \frac{x^3 +3x^2Δx+3x(Δx)^2 + (Δx)^3-x^3}{Δx} $
$ f’(x) = \lim\limits_{ Δx\to 0} \frac{ Δx(3x^2+3xΔx + (Δx)^2)}{Δx} $
$f’(x) = \lim\limits_{ Δx\to 0} (3x^2+3xΔx + (Δx)^2) $
$ f’(x) = 3x^2+3x(0) + 0^2 $
$ f’(x) = 3x^2 $
When x = a, $f(0) = 0$ and $f’(0) = 0 $
Find the equation for the tangent line:
$(y - y_1) = m(x-x_1) $
$(y - 0)=0(x-0) $
$y_{tangent} =0 $
$f’(x) = 3x^2 $
The equation of the tangent line: $y = 0 $
