Question bank — Implicit differentiation — technique, applications

True — Chain Rule ke according, $y$ wali kisi bhi cheez ko $x$ ke w.r.t. differentiate karne par "toll" $\frac{dy}{dx}$ milta hai, kyunki $y$ ki $x$ ke respect mein inside-rate-of-change $\frac{dy}{dx}$ hai, $1$ nahi.

False — $x^2+y^2=25$ do branches define karta hai (upper aur lower semicircle); implicit differentiation quietly us whichever branch ko handle karta hai jis par tumhara point hai, kyunki $-x/y$ mein $y$ ke sign se correct sign aa jaata hai.

False — kyunki dono methods same curve describe karte hain, toh har point par unhe same slope dena hi chahiye; implicit answer bas mixed $x,y$ form mein likha hota hai. Circle ki top branch $y=\sqrt{25-x^2}$ par, $-x/y$ mein substitute karne par exactly $-x/\sqrt{25-x^2}$ milta hai — wohi explicit derivative — toh dono line-for-line agree karte hain.

False — jahan $F_y=0$ hoga wahan fail ho jaata hai (division by zero); precisely wahi pe tangent vertical hoti hai aur Implicit Function Theorem $y$ ko $x$ ka function guarantee nahi karta.

True — dono sides $x$ ki equal functions hain, toh unke derivatives bhi equal hain; isi wajah se hum dono sides ko ek saath differentiate kar sakte hain, constants aur sab kuch.

False — $(5,0)$ par $y=0$ se divide karna padega; tangent vertical hai (undefined slope), jo formula blowing up karke dikhata hai, koi number dene ki jagah.

False — Logarithmic Differentiation is implicit differentiation hai jo $\ln y=(\text{stuff})$ par apply hota hai; left side par $\frac{1}{y}\frac{dy}{dx}$ sirf $\ln y$ par chain-rule toll hai.

False — $y^{\prime \prime }$ ko sirf position $(x,y)$ ka function hona chahiye, taaki tum ek point plug karke number pa sako; lekin $y^{\prime }$ koi independent quantity nahi hai — woh khud known expression $-x/y$ hai. $y^{\prime }$ ko chhodna matlab tumhara answer secretly abhi bhi ek unevaluated derivative contain karta hai, toh woh $(x,y)$ ka usable function nahi hai abhi — use substitute karna hi use ek banata hai.

Chain-rule toll missing hai; $y$ ek function of $x$ hai, toh $\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$ hoga, $2y$ nahi. Ye sirf isliye $\frac{d}{dx}(x^2)=2x$ jaisa lagta hai kyunki dono squares hain.

$xy$ do functions ka product hai, toh Product Rule lagega: $\frac{d}{dx}(xy)=1\cdot y+x\cdot \frac{dy}{dx}=y+x{y}^{\prime }$. $xy$ ko "ek lump" maanna dono terms ko throw away kar deta hai.

Constant ki rate of change zero hoti hai, toh $\frac{d}{dx}(25)=0$; equation ka balance exactly isliye preserve hota hai kyunki constants vanish ho jaate hain.

Tum $\sin (xy)=x+y$ ko closed form mein $y$ ke liye solve nahi kar sakte, toh differentiate karne ke liye koi explicit cheez hai hi nahi; implicit differentiation hi ek available route hai.

Outer chain rule sirf aadha hua hai — tumhe $\frac{d}{dx}(xy)=y+x{y}^{\prime }$ se multiply karna hoga, jo deta hai $\cos (xy)(y+x{y}^{\prime })$. Inside $x$ ka function hai, toh uska derivative $1$ nahi hai.

$\arcsin$ ki range, yaani $y\in [-\frac{\pi }{2},\frac{\pi }{2}]$, par $\cos y\ge 0$ hota hai, toh sirf positive root correct hai: $y^{\prime }=\frac{1}{\sqrt{1-x^2}}$, domain $x\in (-1,1)$ par valid (dekho Derivatives of Inverse Functions).

Power rule ko constant exponent chahiye; yahan exponent variable $x$ hai, toh rule apply nahi hota. Pehle log lo: $\ln y=x\ln x$, jisse $y^{\prime }=x^x(\ln x+1)$ milta hai.

Tum point pehle plug in kar sakte ho, lekin $y^{\prime }$ isolate karna abhi bhi zaroori hai: $y^{\prime }(27-18)=18-27$ se $y^{\prime }=-1$ milta hai; isolation skip karna $y^{\prime }$ undetermined chhod deta hai.

Kyunki hum $x$ ke respect mein differentiate kar rahe hain: $x$ apne aap ke saath rate $1$ par change hota hai ($\frac{dx}{dx}=1$), jabki $y$ rate $\frac{dy}{dx}$ par change hota hai, jo ek unknown hai jise hum solve kar rahe hain.

Hum curve ko $y=+\sqrt{\cdots }$ aur $y=-\sqrt{\cdots }$ mein split nahi karte; ek single formula $-x/y$ automatically point ki $y$-value ke sign se sahi sign carry karta hai, dono branches ek saath cover karta hai (neeche branch figure dekho).

$F_y=0$ matlab curve locally vertical hai, toh $y$ ko wahan $x$ ka function nahi likha ja sakta (Implicit Function Theorem); zero se division algebra ka warning hai ki vertical tangent hai.

$\ln$ exponents ko multipliers mein aur products ko sums mein convert karta hai ($\ln (x^x)=x\ln x$), ek aisi case ko jo koi elementary rule cover nahi karta, ek aise product mein badal deta hai jo Product Rule handle kar sakta hai.

Hum formula derive karte hain: $\sin y=x$ set karke aur implicitly differentiate karne se woh general machine milti hai jo har inverse-function derivative produce karta hai — result yaad karna skip kar deta hai ye samajhna ki $x\in (-1,1)$ par woh $1/\sqrt{1-x^2}$ kyun hai.

Related Rates mein dono $x$ aur $y$ time ke functions hain, toh hum $t$ ke w.r.t. differentiate karte hain; phir har variable toll deta hai ($\frac{dx}{dt}$ ya $\frac{dy}{dt}$) — same chain-rule logic, alag independent variable.

Normal tangent ke perpendicular hoti hai, toh uska slope $-1/m$ hai (dekho Tangent and Normal Lines); implicit differentiation $m=\frac{dy}{dx}$ supply karta hai chahe $y=f(x)$ nahi likhi ja sake.

Woh blow up ho jaata hai (denominator $y\to 0$), jo correct signal hai ki slope undefined/vertical hai, koi mistake nahi — geometry aur algebra agree karte hain (upar vertical-tangent figure dekho).

$0/0$ milega, ek indeterminate form jo ek singular point flag karta hai (jaise folium of Descartes ka origin par crossing), jahan ek se zyaada tangents exist kar sakti hain aur simple formula fail ho jaata hai.

$\ln x$ aur $x^x$ sirf $x>0$ ke liye real-valued hain (usual sense mein); non-positive base ka $\ln$ undefined hai, toh logarithmic method us domain tak restricted hai.

$y=0$ par (points $(\pm 5,0)$) tangent vertical hai aur $y$ locally $x$ ka function nahi hai, toh $y^{\prime }$ aur $y^{\prime \prime }$ dono wahan undefined hain — denominator mein $y^3$ ise explicit kar deta hai.

Steps formally $\frac{dy}{dx}=-x/y$ tak chal jaate hain, lekin equation ke koi real solutions nahi hain, toh koi real curve nahi aur hence koi real tangent nahi — derivative trust karne se pehle hamesha check karo ki locus nonempty hai.

Slope exactly $0$ hoga — horizontal tangent — jo perfectly valid hai; sirf zero denominator ($F_y=0$) trouble karta hai, zero numerator nahi.