Differentiation

$f^{\mathrm{\prime }}(x)=3x^2-6x=3x(x-2)$. Roots: $x=0$ and $x=2$ — two critical points.

By the power rule, $\frac{d}{dx}{x}^3=3x^2$.

Differentiating term by term: $(\sin x{)}^{\mathrm{\prime }}+(\cos x{)}^{\mathrm{\prime }}=\cos x-\sin x$.

$f^{\mathrm{\prime }}(x)=3(x^2+1{)}^2\cdot 2x=6x(x^2+1{)}^2$. At $x=1$: $6\cdot 1\cdot 4=24$.

${\displaystyle \frac{f(3+h)-f(3)}{h}}={\displaystyle \frac{(3+h{)}^2-9}{h}}={\displaystyle \frac{9+6h+h^2-9}{h}}={\displaystyle \frac{6h+h^2}{h}}=6+h$.

${\displaystyle \underset{h\to 0}{\lim }(6+h)=6}$.

$f^{\mathrm{\prime }}(x)=2x-6$.

Setting $f^{\mathrm{\prime }}(c)=0$: $2c-6=0⟹c=3$.

Since $f^{\mathrm{\prime }\mathrm{\prime }}(x)=2>0$, this is indeed a minimum.

Average rate: ${\displaystyle \frac{f(4)-f(0)}{4-0}}={\displaystyle \frac{16-0}{4}}=4$.

$f^{\mathrm{\prime }}(x)=2x$, so $f^{\mathrm{\prime }}(c)=4⟹2c=4⟹c=2$.

$2\in (0,4)$ ✓.

$f$ is a polynomial (continuous on $[-2,2]$, differentiable on $(-2,2)$), and $f(-2)=0=f(2)$, so Rolle's hypotheses hold.

$f^{\mathrm{\prime }}(x)=2x$, so $f^{\mathrm{\prime }}(c)=0⟹c=0$.

$\frac{d}{dx}\ln u=\frac{u^{\mathrm{\prime }}}{u}$. Here $u^{\mathrm{\prime }}=2x$, so $f^{\mathrm{\prime }}(x)={\displaystyle \frac{2x}{x^2+1}}$.

$f^{\mathrm{\prime }}(x)=5(x^2+1{)}^4\cdot 2x=10x(x^2+1{)}^4$.