Exercises — Higher-order derivatives — notation, physical meaning

WHAT: power rule $\frac{d}{dx}{x}^n=n{x}^{n-1}$ ko teen baar apply karo. WHY power rule: plain power of $x$ ke liye yeh sabse fast aur sahi tool hai — yeh "$x^n$ ka slope kya hai?" ka jawab deta hai bina limits par wapas jaaye.

• $f^{\prime }(x)=5x^4$ — $5$ ko neeche laao, exponent ko ek se ghatao.
• $f^{\prime \prime }(x)=5\cdot 4x^3=20x^3$ — $5x^4$ ko differentiate karo.
• $f^{\prime \prime \prime }(x)=20\cdot 3x^2=60x^2$ — $20x^3$ ko differentiate karo.

WHAT this asks: bookkeeping ko decode karo, kuch compute nahi karna.

• (a) $\frac{dy}{dx}$ = "$\frac{d}{dx}$ ko $y$ par ek baar apply karo" = first derivative = $f^{\prime }(x)$.
• (b) $\frac{d^{3}y}{d{x}^3}$ = ${\left(\frac{d}{dx}\right)}^{3}y$ = "$\frac{d}{dx}$ ko teen baar apply karo" = $f^{\prime \prime \prime }(x)$. Upar ke $d$'s ginne jaate hain; neeche ke $dx$'s ginne jaate hain — ye operator tallies hain, powers nahi.

WHY time differentiate karte hain: velocity hai kitni tezi se position badlti hai, isliye yeh $\frac{ds}{dt}$ hai; acceleration hai kitni tezi se velocity badlti hai, isliye yeh $\frac{dv}{dt}=\frac{d^{2}s}{d{t}^2}$ hai (dekho Kinematics — position, velocity, acceleration).

• $v(t)=s^{\prime }(t)=8t$ — first derivative.
• $a(t)=v^{\prime }(t)=8$ — $s$ ka second derivative; yahan yeh constant $8{\text{m/s}}^2$ hai.

• $v(t)=s^{\prime }(t)=20-10t$.
• $a(t)=v^{\prime }(t)=-10{\text{m/s}}^2$ (constant downward — gravity). Highest point: ball momentarily ruk jaati hai, toh $v=0\Rightarrow 20-10t=0\Rightarrow t=2$ s. WHY yahan $v=0$: top par velocity upar (+) se neeche (−) switch karti hai, isliye zero se guzarti hai. Max height: $s(2)=20(2)-5(2{)}^2=40-20=20$ m.

Step 1 — WHY ek phase shift ek derivative ke barabar hai. Ek key trig identity hai: sine ke angle mein $\frac{\pi }{2}$ jodhne se woh cosine ban jaata hai, $\sin \left(x+\frac{\pi }{2}\right)=\cos x$. Notice karo ki $\cos x$ exactly $\frac{d}{dx}\sin x$ hai. Toh $\sin$ ka ek differentiation uske angle ko $+\frac{\pi }{2}$ shift karne ke barabar hai. Step 2 — repeat karo aur shifts gino. Kyunki $\sin$ ko differentiate karne par hamesha ek sine-ya-cosine milti hai, wahi identity har baar apply hoti hai. $n$ baar differentiate karne par angle mein kul $n$ baar $\frac{\pi }{2}$ jud jaata hai: $f^{(n)}(x)=\sin \left(x+n\cdot \frac{\pi }{2}\right).$ Step 3 — sanity checks. $n=1$: $\sin \left(x+\frac{\pi }{2}\right)=\cos x$ ✓ (matches $\frac{d}{dx}\sin x$). $n=2$: $\sin (x+\pi )=-\sin x$ ✓. Isliye derivatives period 4 ke saath cycle karte hain ($2\pi$ jodhne par start par wapas aate ho) — dekho Trigonometric derivatives. Ab $n=7$: $f^{(7)}(x)=\sin \left(x+\frac{7\pi }{2}\right)$. Kyunki $\frac{7\pi }{2}=4\pi -\frac{\pi }{2}$, yeh $\sin \left(x-\frac{\pi }{2}\right)=-\cos x$ ke barabar hai. Direct check: $\sin \to \cos \to -\sin \to -\cos \to \sin \to \cos \to -\sin \to -\cos$. 7th arrow $-\cos x$ par land karta hai. ✓

WHY har baar ek factor bahar aata hai: chain rule se $\frac{d}{dx}{e}^{2x}=2e^{2x}$; har differentiation ek aur $2$ se multiply karta hai. $g^{\prime }(x)=2e^{2x},g^{\prime \prime }(x)=4e^{2x},\dots g^{(n)}(x)=2^n{e}^{2x}$. $g^{(4)}(0)=2^4{e}^0=16$.

• $f^{\prime }(x)=3x^2-3$, phir $f^{\prime \prime }(x)=6x$. WHY $f^{\prime \prime }$ ka sign matter karta hai: $f^{\prime \prime }$ woh rate hai jis par slope change hota hai, yaani 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 $⌣$.
• $x=0$: $f^{\prime \prime }=0$ aur sign flip hota hai → inflection point at $(0,0)$.

Neeche figure — kya dhundna hai: curve ek baar navy mein draw ki gayi hai, phir bending visible banane ke liye do colours mein re-trace ki gayi hai. Magenta left arm ($x<0$) downward cup karta hai; orange right arm ($x>0$) upward cup karta hai. Violet dot origin par woh jagah hai jahan dono bends milte hain — inflection point jahan $f^{\prime \prime }=6x$ zero cross karke sign flip karta hai.

WHAT rule sach mein kehta hai: ek inflection point ke liye $f^{\prime \prime }$ ka sign change hona zaroori hai, sirf zero ko touch karna kaafi nahi. $h^{\prime \prime }(x)=12x^2$, $0$ ke dono taraf $\ge 0$ hai — yeh zero tak gira, lekin kabhi negative nahi gaya. Toh concavity poore time "up" rehti hai. Conclusion: $x=0$ ek inflection point nahi hai (yeh ek flat-bottomed minimum hai). $f^{\prime \prime }=0$ necessary hai lekin sufficient nahi.

Neeche figure — kya dhundna hai: navy $x^4$ curve origin par violet dot ke dono taraf cup-shaped ($⌣$) rehti hai. Haalaanki curve wahan momentarily flat ho jaati hai ($h^{\prime \prime }(0)=0$), yeh kabhi $⌢$ bend ki taraf nahi palti — toh koi sign change nahi aur koi inflection point nahi. Isse pichle figure se contrast karo, jahan do colours genuinely alag bends the.

• $v(t)=3t^2-12t+9=3(t-1)(t-3)$.
• $a(t)=6t-12=6(t-2)$. WHY signs compare karte hain: speed tab badhti hai jab velocity aur acceleration same direction mein hon (unka product $v\cdot a>0$); agar oppose karein, toh object slow ho raha hai. $t\ge 0$ par sign chart: | 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 | Toh yeh $(1,2)\cup (3,\infty )$ par speed up karta hai.

• $v(t)=s^{\prime }(t)=\cos t$ — WHY: velocity position ka first derivative hai.
• $a(t)=v^{\prime }(t)=-\sin t$ — WHY: acceleration velocity ka derivative hai, $\frac{d}{dt}\cos t=-\sin t$.
• $j(t)=a^{\prime }(t)=-\cos t$ — WHY: upar di gayi definition se, jerk $j=a^{\prime }$ acceleration ka derivative hai, aur $\frac{d}{dt}(-\sin t)=-\cos t$. Observation: $a(t)=-\sin t=-s(t)$. WHY yeh special hai: acceleration hamesha origin ki taraf waapas point karta hai aur displacement ke proportional hai — yeh simple harmonic motion ki pehchaan hai (ek spring par mass). Negative sign woh restoring force hai jo use ghar kheechtaa hai.

WHY power rule abhi bhi apply hota hai: $x^{-1}$ ek power hai, toh $\frac{d}{dx}{x}^k=k{x}^{k-1}$ negative $k$ ke liye bhi kaam karta hai.

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

Cosine ka cycle: $g=\cos x,g^{\prime }=-\sin x,g^{\prime \prime }=-\cos x,g^{\prime \prime \prime }=\sin x,g^{(4)}=\cos x$. $x=0$ par: $g(0)=1,g^{\prime }(0)=0,g^{\prime \prime }(0)=-1,g^{\prime \prime \prime }(0)=0,g^{(4)}(0)=1$. Pattern: even orders $+1,-1,+1,\dots$ alternate karte hain; saare odd orders zero hain. Isliye $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots$ (sirf even powers bachte hain).

WHY degree har baar ek se ghatti hai: $x^m$ ko differentiate karne par $m{x}^{m-1}$ milta hai — ek degree kam. Degree $6$ se start karke: $6$ derivatives ke baad top term ek constant hai; 7th derivative use khatam kar deta hai.

• $p^{(7)}(x)=0$ sab $x$ ke liye.
• Sabse chhota $k$ jiske liye $p^{(k)}\equiv 0$ ho, woh hai $k=7=\mathrm{deg}p+1$.

WHY product rule: $x^2{e}^x$ do functions ka product hai, toh $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$ (Power Rule and differentiation rules).

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

WHAT "un-differentiate" ka matlab hai: ek aisi function dhundho jiska derivative diya gaya wala hai (kinematic ladder ka reverse).

• $v(t)=gt+v_0$ — kyunki $\frac{d}{dt}(gt+v_0)=g$.
• $s(t)=\frac{1}{2}g{t}^2+v_{0}t+s_0$ — kyunki iska derivative $gt+v_0=v(t)$ hai. Numbers: $s(2)=\frac{1}{2}(10)(2^2)+0\cdot 2+100=20+100=120$ m. Yeh jaana-pehchana $s=\frac{1}{2}g{t}^2+v_{0}t+s_0$ hai — sirf acceleration ko do baar integrate karke recover kiya gaya.