Exercises — Derivatives of all six trig functions

L1.1 Upar diye sum picture ke hisaab se, term by term differentiate karo: $f^{\prime }(x)=\cos x+(-\sin x)=\cos x-\sin x.$ Concretely (figure s01 dekho): $\sin x$ ka slope wave $\cos x$ hai, aur $\cos x$ ka slope wave $-\sin x$ hai (wohi wave ulti, kyunki $\cos$ wahan girta hai jahan $\sin$ chadhhta hai). Un do slope waves ko add karo aur $f^{\prime }$ milta hai.

L1.2 Vertical-stretch picture ke hisaab se, constant $4$ slope ko multiply karta hai: $g^{\prime }(x)=4{\sec }^{2}x.$

L1.3 Term by term, ${\sec }^{\prime }=\sec x\tan x$ aur ${\csc }^{\prime }=-\csc x\cot x$ use karte hue (upar re-derive kiya): $\frac{d}{dx}(\sec x-\csc x)=\sec x\tan x-(-\csc x\cot x)=\sec x\tan x+\csc x\cot x.$ KYA hua yahan: $\csc x$ ke aage ka minus, ${\csc }^{\prime }$ ke built-in minus se mila, aur do negatives ne ek plus banaya.

L2.1 — Product Rule $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$, $u=x^3,v=\cos x$ ke saath. Rectangle ki do sides badhti hain: $u^{\prime }=3x^2,v^{\prime }=-\sin x.$ $f^{\prime }(x)=3x^2\cos x+x^3(-\sin x)=3x^2\cos x-x^3\sin x.$ Dhyan do doosra term negative hai: jaise $x$ badhta hai, $\cos x$ (figure s01 se) neeche aa raha hota hai, toh rectangle ka woh side shrink ho raha hai — minus picture hai, sirf algebra nahi.

L2.2 — Chain Rule. Outer function $\tan (\square )$ se ${\sec }^2(\square )$ milta hai; inner $\square =5x$ se $5$ milta hai: $g^{\prime }(x)={\sec }^2(5x)\cdot 5=5{\sec }^2(5x).$ WHY extra $5$ hai: chain rule kehta hai "outside ka derivative $\times$ inside ka derivative." Inside ka factor bhool jaana classic error hai.

L2.3 — Quotient Rule ${\left(\frac{u}{v}\right)}^{\prime }=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$, $u=\sin x$ ($u^{\prime }=\cos x$), $v=x^2$ ($v^{\prime }=2x$) ke saath. Slope-of-a-ratio picture use karte hue: $\sin$ ka badhna ratio ko upar push karta hai ($+\cos x\cdot x^2$), jabki $x^2$ ka badhna ise neeche push karta hai ($-\sin x\cdot 2x$): $h^{\prime }(x)=\frac{\cos x\cdot x^2-\sin x\cdot 2x}{(x^2{)}^2}=\frac{x^2\cos x-2x\sin x}{x^4}.$ Ek $x$ cancel karo ($x\ne 0$ ke liye valid): $h^{\prime }(x)=\frac{x\cos x-2\sin x}{x^3}.$

L3.1 $y^{\prime }={\sec }^{2}x=\frac{1}{{\cos }^{2}x}$. $x=\pi /4$ par, $\cos (\pi /4)=\frac{\sqrt{2}}{2}$, toh ${\cos }^2(\pi /4)=\frac{1}{2}$: $y^{\prime }{|}_{\pi /4}=\frac{1}{1/2}=2.$ Interpretation: $\tan$ ka graph wahan slope $2$ se chadh raha hai — ek $45^{\circ }$ line se do guna steep. Figure s04 mein orange dot $(\pi /4,1)$ par baith hai aur usse hone wali dashed violet line ka slope $2$ hai; jaise $x$ dotted asymptote $x=\pi /2$ ki taraf badhta hai, $\cos x\to 0$ toh ${\sec }^{2}x\to \infty$ aur curve rocket ki tarah upar jaata hai.

L3.2 Horizontal tangent ka matlab slope $=0$. $y=\sin x$ ke liye, $y^{\prime }=\cos x$, toh hum solve karte hain $\cos x=0⟹x=\frac{\pi }{2},\frac{3\pi }{2}.$ Ye exactly sine wave ki crest aur trough hain — jahan woh momentarily uthna ya girna band kar deti hai. Figure s01 dobara dekho: $x=\pi /2$ par magenta wave flat hai aur violet slope-wave zero cross karti hai.

L3.3 Point: $x=\pi /2$ par, $y=\cos (\pi /2)=0$, toh point hai $\left(\frac{\pi }{2},0\right)$. Slope: $y^{\prime }=-\sin x$, toh $y^{\prime }{|}_{\pi /2}=-\sin (\pi /2)=-1$. $\left(\frac{\pi }{2},0\right)$ se guzarne wali tangent line slope $-1$ ke saath: $y-0=-1\left(x-\frac{\pi }{2}\right)⟹y=-x+\frac{\pi }{2}.$

L4.1 — Bahar Product Rule (do-sided rectangle), $\sec (3x)$ factor par Chain Rule. Maano $u=x^2$ ($u^{\prime }=2x$) aur $v=\sec (3x)$. $v^{\prime }$ ke liye: $v^{\prime }=\sec (3x)\tan (3x)\cdot 3=3\sec (3x)\tan (3x).$ Ab $f^{\prime }=u^{\prime }v+u{v}^{\prime }$ assemble karo (side-strip plus top-strip): $f^{\prime }(x)=2x\sec (3x)+x^2\cdot 3\sec (3x)\tan (3x)=2x\sec (3x)+3x^2\sec (3x)\tan (3x).$ Shared $\sec (3x)$ factor karo: $f^{\prime }(x)=\sec (3x)[2x+3x^2\tan (3x)].$

L4.2 — Quotient Rule, $u=1+\sin x$ ($u^{\prime }=\cos x$), $v=\cos x$ ($v^{\prime }=-\sin x$) ke saath. Slope-of-a-ratio picture ke hisaab se, numerator $u$ ka badhna $\cos x\cdot \cos x$ add karta hai aur denominator $v$ ka girna (dhyan do $v^{\prime }=-\sin x$) $(1+\sin x)(-\sin x)$ contribute karta hai: $g^{\prime }(x)=\frac{\cos x\cdot \cos x-(1+\sin x)(-\sin x)}{{\cos }^{2}x}=\frac{{\cos }^{2}x+\sin x+{\sin }^{2}x}{{\cos }^{2}x}.$ Pythagorean identity sin^2 + cos^2 = 1 use karo: numerator ${\cos }^{2}x+{\sin }^{2}x=1$, toh bacha $g^{\prime }(x)=\frac{1+\sin x}{{\cos }^{2}x}.$ WHY ye beautiful hai: identity teen mein se do terms ko ek clean $1$ mein collapse kar deti hai. Domain caveat: ye result sirf wahan hold karta hai jahan $\cos x\ne 0$ — yaani vertical asymptotes $x=\frac{\pi }{2}+k\pi$ ($k$ koi bhi integer) se door. Un points par original $g(x)$ khud blow up ho jaata hai, toh na $g$ na $g^{\prime }$ wahan defined hai. Quotient derivative quote karte waqt kabhi apne denominator ke zeros mat bhulo.

L4.3 — Chain Rule. Outer $\sin (\square )$ se $\cos (\square )$ milta hai; inner $\square =\cos x$ se $-\sin x$ milta hai: $y^{\prime }=\cos (\cos x)\cdot (-\sin x)=-\sin x\cos (\cos x).$ DHYAN RAHO: $\cos (\cos x)$ ka matlab hai "cosine of the number $\cos x$" — ise mat simplify karo ${\cos }^{2}x$ mein. Ye do alag cheezein hain.

L5.1 Velocity, position ka derivative hai (Chain Rule, inner $2t$ se $2$ milta hai): $v(t)=\frac{d}{dt}[3\cos (2t)]=3\cdot (-\sin (2t))\cdot 2=-6\sin (2t)\text{m/s}.$ Acceleration, velocity ka derivative hai (phir chain rule, inner se $2$ milta hai): $a(t)=\frac{d}{dt}[-6\sin (2t)]=-6\cos (2t)\cdot 2=-12\cos (2t){\text{m/s}}^2.$ Ab $x(t)=3\cos (2t)$ se compare karo: $x$ ko $-4$ se multiply karo: $-4x(t)=-12\cos (2t)=a(t).✓$ Toh $a=-4x$, jo $a=-{\omega }^{2}x$ hai jahan ${\omega }^2=4$, yaani $\boxed{\omega =2}$ rad/s. WHY $\omega$ inner derivative hai (figure s05): position $3\cos (2t)$, radius $3$ ke ek circle par spin karte point ka shadow (horizontal projection) hai. Ek second mein woh point angle $2t$ sweep karta hai, toh woh $2$ radians per second spin karta hai — wahi spin rate $\omega$ hai. Differentiate karna poochta hai "angle kitni tezi se badal raha hai?", aur chain rule ke hisaab se woh rate inner derivative $\frac{d}{dt}(2t)=2$ hai. Do derivatives phasor ko do baar spin karti hain, $2^2=4={\omega }^2$ pull out karte hue. Physics aur chain rule ek hi fact hain.

L5.2 Quotient Rule, $u=\cos x$ ($u^{\prime }=-\sin x$), $v=\sin x$ ($v^{\prime }=\cos x$) ke saath: $\frac{d}{dx}\cot x=\frac{(-\sin x)(\sin x)-(\cos x)(\cos x)}{{\sin }^{2}x}=\frac{-{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{2}x}.$ $-1$ factor out karo aur ${\sin }^{2}x+{\cos }^{2}x=1$ apply karo: $=\frac{-({\sin }^{2}x+{\cos }^{2}x)}{{\sin }^{2}x}=\frac{-1}{{\sin }^{2}x}=-{\csc }^{2}x.$ Kahan undefined hai? $\cot x$ aur uska derivative dono wahan blow up karte hain jahan $\sin x=0$, yaani $x=0,\pm \pi ,\pm 2\pi ,\dots$ — denominator wahan vanish ho jaata hai, toh koi slope exist nahi karta (vertical asymptotes).

L5.3 Point: $x=\pi /3$ par, $\cos (\pi /3)=\frac{1}{2}$, toh $\sec (\pi /3)=\frac{1}{1/2}=2$. Point hai $\left(\frac{\pi }{3},2\right)$. Slope: $y^{\prime }=\sec x\tan x$. $\sec (\pi /3)=2$ aur $\tan (\pi /3)=\sqrt{3}$ ke saath: $y^{\prime }{|}_{\pi /3}=2\sqrt{3}.$ $\left(\frac{\pi }{3},2\right)$ se guzarne wali tangent line slope $2\sqrt{3}$ ke saath: $y-2=2\sqrt{3}\left(x-\frac{\pi }{3}\right).$

Answers chhupa lo. Kya tum memory se bata sakte ho: (a) "co" functions minus kyun carry karte hain, (b) $\sin x$ kahan flat hai, aur (c) $\cos (2t)$ ko do baar differentiate karne par factor $4$ kyun aata hai? (a) Ye $\sin ,\tan ,\sec$ ke swapped-aur-negated partners hain. ::: "All Co's are Negative." (b) Jahan $\cos x=0$, yaani $x=\frac{\pi }{2},\frac{3\pi }{2}$ (crest aur trough). ::: Slope $=\cos x$. (c) Har derivative chain rule se inner derivative $2$ pull out karta hai; $2\times 2=4$. ::: Woh $4={\omega }^2$ hai.