Comprehensive Exercises

$f'(x) = 2x + 2$, so $f'(1) = 4$. (Using the limit definition yields the same result.)

$\lim_{x\to 0}|x| = 0 = f(0)$ ⇒ continuous.
Left derivative $=-1$, right derivative $=1$ ⇒ not differentiable.

Product rule: $f'(x) = 3x^2\sin x + x^3\cos x$.

Quotient rule: $f'(x) = \dfrac{x^2 + 2x - 1}{(x+1)^2}$.

Chain rule (three layers): $f'(x) = 2x\,e^{x^2}\cos(e^{x^2})$.

Let $y=\arcsin(x^2)$, so $x^2=\sin y$. Then $2x = \cos y\;y'$ and $y' = \dfrac{2x}{\sqrt{1 - x^4}}$.

$3x^2 + 3y^2y' = 3y + 3xy'$ ⇒ $y' = \dfrac{y - x^2}{y^2 - x}$.

$\dfrac{dy}{dx} = \dfrac{3t^2}{2t} = \dfrac{3t}{2}$, and $\dfrac{d^2y}{dx^2} = \dfrac{3}{4t}$.

Directly by Rolle’s theorem.

MVT on $f(t)=\sin t$ gives $\cos \xi = \dfrac{\sin y - \sin x}{y-x}$ with $|\cos \xi|\le 1$.

$\tfrac{0}{0}$ type → $\lim \dfrac{\cos x}{1} = 1$.

Two rounds: $\dfrac{2x}{e^x} \to \dfrac{2}{e^x} \to 0$.

Rewrite $\dfrac{\ln x}{1/x}$ ($\tfrac{\infty}{\infty}$). Derivatives give $-x \to 0$.

$f'(x)=3x(x-2)$ ⇒ critical points $0,2$.
f''(0)<0 ⇒ local max at $(0,2)$.
$f''(2)>0$ ⇒ local min at $(2,-2)$.

$f'(1)=1$ ⇒ $y-0 = 1(x-1)$ ⇒ $y = x-1$.

Check $x=0,2,3$: $f(0)=2$, $f(2)=-2$, $f(3)=2$. Max $2$, min $-2$.

Log-differentiate: $\ln y = x\ln x$ ⇒ $\dfrac{y'}{y} = \ln x + 1$ ⇒ $y' = x^x(\ln x + 1)$.

$f''(x)=6(x-1)$.
Concave down on $(-\infty,1)$, up on $(1,\infty)$.
Inflection at $(1,0)$.

$\dfrac{2\xi}{3\xi^2} = \dfrac{3}{7}$ ⇒ $\xi = \dfrac{14}{9}$.

$f'(1)=2,\ f''(1)=2$. Curvature $\kappa = \dfrac{|2|}{[1+2^2]^{3/2}} = \dfrac{2}{5\sqrt{5}}$.