Answer
$\frac {dy}{dx} = 4x^3 $
Work Step by Step
$ y = x^4 $
$\frac {dy}{dx} = \lim\limits_{∆x \to 0} \frac {f(x+∆x)-f(x)}{∆x} $
$\frac {dy}{dx} = \lim\limits_{∆x \to 0} \frac {(x+∆x)^4 -x^4}{∆x} $
$\frac {dy}{dx} = \lim\limits_{∆x \to 0} \frac {x^4 + 4x^3∆x+6x^2∆x^2+4x∆x^3+∆x^4-x^4}{∆x} $
$\frac {dy}{dx} = \lim\limits_{∆x \to 0} \frac {∆x(4x^3+6x^2∆x+4x∆x^2+∆x^3)}{∆x} $
$\frac {dy}{dx} = \lim\limits_{∆x \to 0} (4x^3+6x^2∆x+4x∆x^2+∆x^3) $
$\frac{dy}{dx} = 4x^3+6x^2(0)+4x^3(0)^2+(0)^3 $
$\frac{dy}{dx} = 4x^3 $
