Answer
$$y'=x^3$$
The graphs of $f(x)$ (red) and $f'(x)$ (green) are plotted below.
Work Step by Step
$$y=\frac{x^4}{4}$$
a) Find the derivative:
$$y'=\lim_{z\to x}\frac{f(z)-f(x)}{z-x}=\lim_{z\to x}\frac{\frac{z^4}{4}-\frac{x^4}{4}}{z-x}=\lim_{z\to x}\frac{\frac{z^4-x^4}{4}}{z-x}$$ $$y'=\lim_{z\to x}\frac{(z^2-x^2)(z^2+x^2)}{4(z-x)}=\lim_{z\to x}\frac{(z-x)(z+x)(z^2+x^2)}{4(z-x)}$$ $$y'=\lim_{z\to x}\frac{(z+x)(z^2+x^2)}{4}$$ $$y'=\frac{(x+x)(x^2+x^2)}{4}=\frac{2x\times2x^2}{4}=x^3$$
b) The graphs are plotted below. On the left side, the red line represents the graph of $y=f(x)$. On the right side, the green line represents the graph of $y=f'(x)$.
c) From the right graph, we see that
- On $(0,\infty)$, $f'(x)$ is positive.
- At $x=0$, $f'(x)$ is zero.
- On $(-\infty,0)$, $f'(x)$ is negative.
d) From the left graph, we see that
- On $(0,\infty)$, $f(x)$ increases as $x$ increases. This corresponds to the interval where $f'$ is positive.
- On $(-\infty,0)$, $f(x)$ decreases as $x$ increases. This corresponds to the interval where $f'$ is negative.
