Exercises — Higher-order derivatives — notation, physical meaning

WHAT: apply the power rule $\frac{d}{dx}{x}^n=n{x}^{n-1}$ three times. WHY the power rule: it is the fastest correct tool for a plain power of $x$ — it answers "what is the slope of $x^n$?" without going back to limits.

• $f^{\prime }(x)=5x^4$ — bring the $5$ down, drop the exponent by one.
• $f^{\prime \prime }(x)=5\cdot 4x^3=20x^3$ — differentiate $5x^4$.
• $f^{\prime \prime \prime }(x)=20\cdot 3x^2=60x^2$ — differentiate $20x^3$.

WHAT this asks: decode the bookkeeping, not compute anything.

• (a) $\frac{dy}{dx}$ = "apply $\frac{d}{dx}$ once to $y$" = the first derivative = $f^{\prime }(x)$.
• (b) $\frac{d^{3}y}{d{x}^3}$ = ${\left(\frac{d}{dx}\right)}^{3}y$ = "apply $\frac{d}{dx}$ three times" = $f^{\prime \prime \prime }(x)$. The top counts the $d$'s; the bottom counts the $dx$'s — they are operator tallies, not powers.

WHY differentiate time: velocity is how fast position changes, so it is $\frac{ds}{dt}$; acceleration is how fast velocity changes, so it is $\frac{dv}{dt}=\frac{d^{2}s}{d{t}^2}$ (see Kinematics — position, velocity, acceleration).

• $v(t)=s^{\prime }(t)=8t$ — first derivative.
• $a(t)=v^{\prime }(t)=8$ — second derivative of $s$; here it is a constant $8{\text{m/s}}^2$.

• $v(t)=s^{\prime }(t)=20-10t$.
• $a(t)=v^{\prime }(t)=-10{\text{m/s}}^2$ (constant downward — gravity). Highest point: the ball is momentarily still, so $v=0\Rightarrow 20-10t=0\Rightarrow t=2$ s. WHY $v=0$ here: at the top the velocity switches from up (+) to down (−), so it passes through zero. Max height: $s(2)=20(2)-5(2{)}^2=40-20=20$ m.

Step 1 — WHY a phase shift equals a derivative. There is one key trig identity: adding $\frac{\pi }{2}$ to the angle of a sine turns it into a cosine, $\sin \left(x+\frac{\pi }{2}\right)=\cos x$. Notice that $\cos x$ is exactly $\frac{d}{dx}\sin x$. So one differentiation of $\sin$ produces the same effect as shifting its angle by $+\frac{\pi }{2}$. Step 2 — repeat and count the shifts. Because differentiating $\sin$ always lands on another sine-or-cosine, the same identity applies again each time. Differentiating $n$ times therefore adds $\frac{\pi }{2}$ to the angle a total of $n$ times: $f^{(n)}(x)=\sin \left(x+n\cdot \frac{\pi }{2}\right).$ Step 3 — sanity checks. $n=1$: $\sin \left(x+\frac{\pi }{2}\right)=\cos x$ ✓ (matches $\frac{d}{dx}\sin x$). $n=2$: $\sin (x+\pi )=-\sin x$ ✓. This is why the derivatives cycle with period 4 (adding $2\pi$ returns you to the start) — see Trigonometric derivatives. Now $n=7$: $f^{(7)}(x)=\sin \left(x+\frac{7\pi }{2}\right)$. Since $\frac{7\pi }{2}=4\pi -\frac{\pi }{2}$, this equals $\sin \left(x-\frac{\pi }{2}\right)=-\cos x$. Direct check: $\sin \to \cos \to -\sin \to -\cos \to \sin \to \cos \to -\sin \to -\cos$. The 7th arrow lands on $-\cos x$. ✓

WHY a factor pops out each time: by the chain rule $\frac{d}{dx}{e}^{2x}=2e^{2x}$; each differentiation multiplies by another $2$. $g^{\prime }(x)=2e^{2x},g^{\prime \prime }(x)=4e^{2x},\dots g^{(n)}(x)=2^n{e}^{2x}$. $g^{(4)}(0)=2^4{e}^0=16$.

• $f^{\prime }(x)=3x^2-3$, then $f^{\prime \prime }(x)=6x$. WHY the sign of $f^{\prime \prime }$ matters: $f^{\prime \prime }$ is the rate at which the slope changes, i.e. the bend (Concavity and the Second-Derivative Test).
• $x<0$: $f^{\prime \prime }=6x<0$ → concave down $⌢$.
• $x>0$: $f^{\prime \prime }=6x>0$ → concave up $⌣$.
• $x=0$: $f^{\prime \prime }=0$ and the sign flips → inflection point at $(0,0)$.

Figure below — what to look for: the curve is drawn once in navy, then re-traced in two colours to make the bending visible. The magenta left arm ($x<0$) cups downward; the orange right arm ($x>0$) cups upward. The violet dot at the origin is where the two bends meet — the inflection point where $f^{\prime \prime }=6x$ crosses zero and flips sign.

WHAT the rule really says: an inflection point needs $f^{\prime \prime }$ to change sign, not merely to touch zero. $h^{\prime \prime }(x)=12x^2$ is $\ge 0$ on both sides of $0$ — it dips to zero but never goes negative. So concavity stays "up" throughout. Conclusion: $x=0$ is not an inflection point (it is a flat-bottomed minimum). $f^{\prime \prime }=0$ is necessary but not sufficient.

Figure below — what to look for: the navy $x^4$ curve stays cup-shaped ($⌣$) on both sides of the violet dot at the origin. Although the curve momentarily flattens there ($h^{\prime \prime }(0)=0$), it never tips over to a $⌢$ bend — so there is no sign change and no inflection point. Contrast this with the previous figure, where the two colours were genuinely different bends.

• $v(t)=3t^2-12t+9=3(t-1)(t-3)$.
• $a(t)=6t-12=6(t-2)$. WHY compare signs: speed increases when velocity and acceleration point the same way (their product $v\cdot a>0$); if they oppose, the object is slowing. Sign chart on $t\ge 0$: | interval | $v$ | $a$ | $v\cdot a$ | speeding up? | |---|---|---|---|---| | $0<t<1$ | $+$ | $-$ | $-$ | no (slowing) | | $1<t<2$ | $-$ | $-$ | $+$ | yes | | $2<t<3$ | $-$ | $+$ | $-$ | no (slowing) | | $t>3$ | $+$ | $+$ | $+$ | yes | So it speeds up on $(1,2)\cup (3,\infty )$.

• $v(t)=s^{\prime }(t)=\cos t$ — WHY: velocity is the first derivative of position.
• $a(t)=v^{\prime }(t)=-\sin t$ — WHY: acceleration is the derivative of velocity, $\frac{d}{dt}\cos t=-\sin t$.
• $j(t)=a^{\prime }(t)=-\cos t$ — WHY: by the definition above, jerk $j=a^{\prime }$ is the derivative of acceleration, and $\frac{d}{dt}(-\sin t)=-\cos t$. Observation: $a(t)=-\sin t=-s(t)$. WHY this is special: the acceleration always points back toward the origin and is proportional to displacement — this is the signature of simple harmonic motion (a mass on a spring). The negative sign is the restoring force pulling it home.

WHY the power rule still applies: $x^{-1}$ is a power, so $\frac{d}{dx}{x}^k=k{x}^{k-1}$ works for negative $k$ too.

• $f^{\prime }(x)=-x^{-2}$
• $f^{\prime \prime }(x)=2x^{-3}$
• $f^{\prime \prime \prime }(x)=-6x^{-4}$ Pattern of signs and factors: $f^{(n)}(x)=(-1{)}^{n}n!x^{-(n+1)}$. Check $n=3$: $(-1{)}^{3}3!x^{-4}=-6x^{-4}$ ✓. $f^{(3)}(1)=-6\cdot 1^{-4}=-6$.

Cycle for cosine: $g=\cos x,g^{\prime }=-\sin x,g^{\prime \prime }=-\cos x,g^{\prime \prime \prime }=\sin x,g^{(4)}=\cos x$. At $x=0$: $g(0)=1,g^{\prime }(0)=0,g^{\prime \prime }(0)=-1,g^{\prime \prime \prime }(0)=0,g^{(4)}(0)=1$. Pattern: even orders alternate $+1,-1,+1,\dots$; all odd orders vanish. This is exactly why $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots$ (only even powers survive).

WHY degree drops by one each time: differentiating $x^m$ gives $m{x}^{m-1}$ — one lower degree. Starting at degree $6$: after 6 derivatives the top term is a constant; the 7th derivative kills it.

• $p^{(7)}(x)=0$ for all $x$.
• Smallest $k$ with $p^{(k)}\equiv 0$ is $k=7=\mathrm{deg}p+1$.

WHY the product rule: $x^2{e}^x$ is a product of two functions, so $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$ (Power Rule and differentiation rules).

• $f^{\prime }(x)=2x{e}^x+x^2{e}^x=(x^2+2x)e^x$.
• Differentiate again: $f^{\prime \prime }(x)=[(2x+2)e^x]+[(x^2+2x)e^x]=(x^2+4x+2)e^x$.
• $f^{\prime \prime }(0)=(0+0+2)e^0=2$.

WHAT "un-differentiate" means: find a function whose derivative is the given one (the reverse of the kinematic ladder).

• $v(t)=gt+v_0$ — because $\frac{d}{dt}(gt+v_0)=g$.
• $s(t)=\frac{1}{2}g{t}^2+v_{0}t+s_0$ — because its derivative is $gt+v_0=v(t)$. Numbers: $s(2)=\frac{1}{2}(10)(2^2)+0\cdot 2+100=20+100=120$ m. This is the familiar $s=\frac{1}{2}g{t}^2+v_{0}t+s_0$ — recovered purely by integrating acceleration twice.