Exercises — Implicit differentiation — technique, applications

$\frac{d}{dx}(x^n)=n{x}^{n-1}$ kyunki andar ($x$) rate $1$ pe change hota hai. Lekin $\frac{d}{dx}(y^n)=n{y}^{n-1}\cdot \frac{dy}{dx}$ kyunki andar ($y$) rate $\frac{dy}{dx}$ pe change hota hai jab $x$ move karta hai. Same rule, extra factor.

Dono sides ko $x$ ke w.r.t. differentiate karo. KYA: har term pe $\frac{d}{dx}$ apply karo. KYU: ek balanced equation balanced rehti hai. $2x+2y\frac{dy}{dx}=0$ $y^2$ term ne toll pay kiya: $\frac{d}{dx}(y^2)=2y\cdot y^{\prime }$. Solve karo: $\frac{dy}{dx}=-\frac{2x}{2y}=-\frac{x}{y}.$

Har term differentiate karo. KYA: $\frac{d}{dx}$ term by term apply karo. KYU toll sirf $y^3$ pe: $x^3$ variable ki plain power hai, toh $\frac{d}{dx}(x^3)=3x^2$ bina toll ke; lekin $y^3$ ke andar $y$ chhupi hai, isliye chain rule $\frac{dy}{dx}$ attach karta hai: $3x^2+3y^2\frac{dy}{dx}=0.$ KYU right side $0$ hai: constant $10$ kabhi change nahi hota, isliye uska derivative $0$ hai. Ab $y^{\prime }$ isolate karo — KYU $3y^2$ se divide karo: kyunki yahi coefficient $y^{\prime }$ ko multiply kar rahi hai: $3y^2\frac{dy}{dx}=-3x^2⟹\frac{dy}{dx}=-\frac{x^2}{y^2}.$

$2y\frac{dy}{dx}=4⟹\frac{dy}{dx}=\frac{4}{2y}=\frac{2}{y}.$ DEKHNE MEIN KAISA LAGTA HAI: vertex $y=0$ pe yeh blow up karta hai — tangent wahan vertical hai, jo rightward opening parabola ki picture se match karta hai.

$xy$ do functions ka product hai. KYU product rule: koi bhi factor constant nahi hai, toh $\frac{d}{dx}(xy)=x^{\prime }y+x{y}^{\prime }$. Upar definition se yaad karo ki $x^{\prime }=\frac{dx}{dx}=1$, toh yeh $1\cdot y+x{y}^{\prime }=y+x\frac{dy}{dx}$ ho jaata hai. $y+x\frac{dy}{dx}=0⟹\frac{dy}{dx}=-\frac{y}{x}.$

$x^{2}y$ aur $y^{2}x$ dono products hain (ek $x$-piece times ek $y$-piece), isliye dono ko product rule chahiye. KYA + KYU $x^{2}y$ pe: socho "(derivative of $x^2$)$\times y$ + $x^2\times$(derivative of $y$)". Pehla piece $\frac{d}{dx}(x^2)=2x$ mein koi toll nahi; doosra piece $\frac{d}{dx}(y)=y^{\prime }$ mein hai. Result: $2xy+x^2{y}^{\prime }$. KYA + KYU $x{y}^2$ pe: "(derivative of $x$)$\times y^2$ + $x\times$(derivative of $y^2$)" $=(1)y^2+x(2y{y}^{\prime })$ — $y^2$ chain rule se toll pay karta hai, akela $x$ nahi. $\underset{\frac{d}{dx}(x^{2}y)}{\underbrace{2xy+x^2{y}^{\prime }}}+\underset{\frac{d}{dx}(x{y}^2)}{\underbrace{y^2+2xy{y}^{\prime }}}=0.$ KYU $y^{\prime }$ terms collect karo: hume $y^{\prime }$ akela chahiye, toh $y^{\prime }$ wali sab cheezein ek side pe aur baaki sab doosri side pe ikkhatti karo: $y^{\prime }(x^2+2xy)=-(2xy+y^2)⟹\frac{dy}{dx}=-\frac{2xy+y^2}{x^2+2xy}.$

$\cos y\frac{dy}{dx}+1=2y\frac{dy}{dx}.$ KYU $\cos y\cdot y^{\prime }$: $\frac{d}{dx}(\sin y)$ = ($\sin$ ka derivative at $y$) $\times$ (rate jis pe $y$ change hota hai) $=\cos y\cdot y^{\prime }$ — chain rule. Collect karo: $\cos y{y}^{\prime }-2y{y}^{\prime }=-1⟹y^{\prime }(\cos y-2y)=-1⟹\frac{dy}{dx}=\frac{1}{2y-\cos y}.$

$3x^2+3y^2{y}^{\prime }=6(y+x{y}^{\prime }).$ KYU: RHS $6xy$ ek product hai $\Rightarrow 6(y+x{y}^{\prime })$. Collect karo: $3y^2{y}^{\prime }-6x{y}^{\prime }=6y-3x^2⟹y^{\prime }=\frac{6y-3x^2}{3y^2-6x}=\frac{2y-x^2}{y^2-2x}.$ $(3,3)$ pe: $y^{\prime }=\frac{2(3)-9}{9-6}=\frac{-3}{3}=-1.$ Tangent line: $y-3=-1(x-3)\Rightarrow y=-x+6$.

L1.1 se, $y^{\prime }=-x/y$.

• Horizontal tangent ka matlab hai $y^{\prime }=0$. Ek fraction tab $0$ hota hai jab uska numerator $0$ ho: $x=0$. Tab $y^2=25\Rightarrow y=\pm 5$. Points: $(0,5)$ aur $(0,-5)$ — circle ke top aur bottom.
• Vertical tangent ka matlab hai $y^{\prime }$ undefined hai, yaani denominator $0$ hai: $y=0$. Tab $x=\pm 5$. Points: $(5,0)$ aur $(-5,0)$ — left aur right edges.

DEKHNE MEIN KAISA LAGTA HAI: circle ke chaar "compass points" — exactly wahan jahan tangent flat aur upright ke beech flip karti hai.

Differentiate karo: $2x+(y+x{y}^{\prime })+2y{y}^{\prime }=0$ ($xy$ pe product rule). $x{y}^{\prime }+2y{y}^{\prime }=-2x-y⟹y^{\prime }(x+2y)=-(2x+y)⟹y^{\prime }=-\frac{2x+y}{x+2y}.$ $(1,2)$ pe: $y^{\prime }=-\frac{2+2}{1+4}=-\frac{4}{5}.$ Tangent: $y-2=-\frac{4}{5}(x-1)\Rightarrow y=-\frac{4}{5}x+\frac{14}{5}.$

$y^{\prime }=-x/y$ se shuru karo. Quotient rule se dobara differentiate karo, yaad rakho $y^{\prime }=-x/y$: $y^{\prime \prime }=-\frac{(1)y-x{y}^{\prime }}{y^2}=-\frac{y-x(-x/y)}{y^2}=-\frac{y+x^2/y}{y^2}=-\frac{y^2+x^2}{y^3}.$ Kyunki $x^2+y^2=25$: $y^{\prime \prime }=-\frac{25}{y^3}.$ $(3,4)$ pe: $y^{\prime \prime }=-\frac{25}{64}.$ KYU $y^{\prime }$ wapas substitute karo: final answer mein $y^{\prime }$ nahi hona chahiye — warna yeh second derivative ka closed formula nahi hai.

KYU pehle log: base $x$ aur exponent $\sin x$ dono vary karte hain — na power rule na exponential rule apply hoti hai. $\ln$ lene se power ek product mein convert ho jaata hai. $\ln y=\sin x\cdot \ln x.$ Implicitly differentiate karo (left side toll pay karta hai, right side ko product rule chahiye): $\frac{1}{y}{y}^{\prime }=\cos x\ln x+\sin x\cdot \frac{1}{x}.$ $y^{\prime }=x^{\sin x}\left(\cos x\ln x+\frac{\sin x}{x}\right).$ $x=\pi /2$ pe: $\sin x=1,\cos x=0$, toh $y=(\pi /2{)}^1=\pi /2$ aur $y^{\prime }=\frac{\pi }{2}\left(0+\frac{1}{\pi /2}\right)=\frac{\pi }{2}\cdot \frac{2}{\pi }=1.$

Maano $y=\arctan x$, toh $\tan y=x$. Implicitly differentiate karo: ${\sec }^{2}y\frac{dy}{dx}=1⟹\frac{dy}{dx}=\frac{1}{{\sec }^{2}y}.$ $x$ mein wapas convert karo: ${\sec }^{2}y=1+{\tan }^{2}y=1+x^2$ (Pythagorean identity). Isliye $\frac{d}{dx}\arctan x=\frac{1}{1+x^2}.$

$\sin$ pe outer chain rule, phir $xy$ pe product rule: $\cos (xy)(y+x{y}^{\prime })=1+y^{\prime }.$ Expand karo aur collect karo: $y\cos (xy)+x\cos (xy)y^{\prime }=1+y^{\prime }⟹y^{\prime }(x\cos (xy)-1)=1-y\cos (xy).$ $\frac{dy}{dx}=\frac{1-y\cos (xy)}{x\cos (xy)-1}.$ $(0,0)$ pe: $\cos (0)=1$, toh $y^{\prime }=\frac{1-0}{0-1}=-1.$ KYU implicit hi ek raasta hai: $\sin (xy)=x+y$ ko closed form mein $y$ ke liye solve nahi kiya ja sakta.

$x$ aur $y$ dono time $t$ pe depend karte hain, toh equation ko $t$ ke w.r.t. differentiate karo (dekho Related Rates): $2x\frac{dx}{dt}+8y\frac{dy}{dt}=0.$ $x=2,y=1,dx/dt=3$ plug karo: $2(2)(3)+8(1)\frac{dy}{dt}=0⟹12+8\frac{dy}{dt}=0⟹\frac{dy}{dt}=-\frac{12}{8}=-\frac{3}{2}.$ Toh $y$ $1.5$ units/s ki rate se decrease ho raha hai. DEKHNE MEIN KAISA LAGTA HAI: ellipse ke upper-right pe rightward move karne ka matlab hai neeche slide karna, toh negative $dy/dt$ bilkul sahi hai.

Differentiate karo: left side $e^y$ toll pay karta hai. $e^y{y}^{\prime }=1+y^{\prime }⟹y^{\prime }(e^y-1)=1⟹\frac{dy}{dx}=\frac{1}{e^y-1}.$ Point check karo: $(1,0)$ pe, $e^0=1$ aur $x+y=1$ ✓. Lekin tab $y^{\prime }=\frac{1}{e^0-1}=\frac{1}{0}$ — undefined! ISKA KYA MATLAB HAI: $(1,0)$ pe tangent vertical hai. Equation ko $F(x,y)=e^y-x-y=0$ likho; tab $F_y=e^y-1$, jo exactly $(1,0)$ pe $0$ hai. Yahi woh case hai jiske baare mein Implicit Function Theorem warn karta hai: jahan $F_y=0$, curve ko differentiable function $y(x)$ ke roop mein nahi likha ja sakta. Formula fail nahi hua — usne vertical tangent detect kiya.