Chapter 1: Further Differentiation — Exercise 2.1
Q1. sec x, csc x, cot x ki derivative derive karo
(a) sec x:
sec x ko 1/cos x likhte hain. Quotient Rule lagate hain (u=1, v=cos x). Numerator mein u’v – uv’ formula use karte hain: (0)(cos x) – (1)(-sin x) = sin x. Denominator cos²x. Simplify karne se sec x tan x milta hai.
Result: d/dx(sec x) = sec x tan x
(b) csc x:
Isi tarah csc x = 1/sin x likh kar Quotient Rule lagate hain. Result mein negative sign aata hai kyunke derivative of sin x = cos x hai (positive), jo negative fraction bana deta hai.
Result: d/dx(csc x) = −csc x cot x
(c) cot x:
cot x = cos x/sin x, Quotient Rule lagate hain. Numerator mein sin²x + cos²x aata hai jo identity se 1 ban jata hai.
Result: d/dx(cot x) = −csc²x
Ye teeno formulas yaad rakhne chahiye, ye baar baar use hote hain aage ke sawalat mein.
Q2. y = x² cos x
Product Rule use karte hain: (first)'(second) + (first)(second)’. x² ki derivative 2x hai, cos x ki derivative -sin x hai.
Result: dy/dx = 2x cos x − x² sin x
Q3. y = x sec x + 1/x
Do parts hain: “x sec x” ke liye Product Rule, aur “1/x” ke liye Power Rule (1/x = x⁻¹, jiski derivative -x⁻² hoti hai).
Product Rule se: sec x + x sec x tan x. Phir -1/x² add hota hai.
Result: dy/dx = sec x(1 + x tan x) − 1/x²
Q4. y = sin t / (1 − cos t)
Quotient Rule lagate hain. Numerator simplify karne par sin²t + cos²t = 1 identity use hoti hai, jo answer ko bohot simple bana deti hai.
Result: dy/dt = −1/(1 − cos t)
Q5. y = (1 + csc t) sin t
Trick: Pehle multiply kar ke simplify karo: sin t + (csc t · sin t). Chunke csc t · sin t = 1 (reciprocal identity), isliye y = sin t + 1 ban jata hai — bohot asaan ho gaya!
Result: dy/dt = cos t
Sabaq: Hamesha pehle dekho ke expression simplify ho sakta hai ya nahi, phir differentiate karo.
Q6. y = sec x + 4√x − 10
Har term ko alag alag differentiate karo. 4√x ko 4x^(1/2) likh kar Power Rule lagayein.
Result: dy/dx = sec x tan x + 2/√x
Q7. y = √[(1+x)/(1−x)]
Chain Rule use hoti hai. u = (1+x)/(1-x) rakh kar y = √u. Pehle u ki derivative Quotient Rule se nikalo, phir Chain Rule se multiply karo.
Result: dy/dx = 1/[(1−x)^(3/2)(1+x)^(1/2)]
Q8. y = (sin x + cos x)^(3/2)
Chain Rule: u = sin x + cos x, y = u^(3/2). Power ki derivative (3/2)u^(1/2) hoti hai, phir u’ (jo cos x – sin x hai) se multiply karte hain.
Result: dy/dx = (3/2)(sin x + cos x)^(1/2)(cos x − sin x)
Q9. y = sec(tan x)
Chain Rule: u = tan x. sec u ki derivative sec u tan u hoti hai, phir u’ (= sec²x) se multiply.
Result: dy/dx = sec(tan x) · tan(tan x) · sec²x
Q10. y = √sin x + cos√x
Do terms hain, dono par alag Chain Rule lagti hai. Pehle term mein sin x andar hai, doosre mein √x andar hai.
Result: dy/dx = cos x/(2√sin x) − sin(√x)/(2√x)
Q11. y = tan²x + cot(x²)
Yahan farq samajhna zaroori hai: tan²x matlab (tan x)², jabke cot(x²) matlab cot of x². Dono par Chain Rule alag tarah lagti hai.
Result: dy/dx = 2 tan x sec²x − 2x csc²(x²)
Q12. y = sin³(cos 2v)
Yahan Chain Rule 2 baar lagti hai (isliye ise “double chain rule” bhi kehte hain): pehle cube ki power, phir sin function, phir cos function, aur aakhir mein 2v ki derivative.
Result: dy/dv = −6 sin²(cos 2v) cos(cos 2v) sin 2v
Q13. Parametric Equations ka dy/dx
(i) x = sin²t, y = cos 2t:
Dono ko t ke sath differentiate karo (dx/dt aur dy/dt), phir dy/dx = (dy/dt)/(dx/dt) formula lagao. Double angle identity (sin 2t = 2 sin t cos t) use karne se answer bohot simple ho jata hai.
Result: dy/dx = −2
(ii) x = √(1+sec t), y = √(tan t):
Isi formula (dy/dt ÷ dx/dt) se solve karte hain, thora lamba calculation hai lekin same method hai.
Result: dy/dx = sec t√(1+sec t) / tan^(3/2) t
Q14. Prove karo: agar x = a sec t, y = b tan t, to dy/dx = b²x/(a²y)
dx/dt aur dy/dt nikal kar dy/dx = (b sec t)/(a tan t) milta hai. Phir sec t = x/a aur tan t = y/b wapas substitute kar ke required form mein le aate hain.
Result: dy/dx = b²x/(a²y) — Proved
Q15. Implicit Differentiation
(i) ax² + by² + 2gx + 2fy + c = 0:
Poori equation ko x ke sath differentiate karo, y ko x ka function samajh kar (isliye y ki derivative ke sath dy/dx bhi likhna padta hai). Phir dy/dx ko alag kar ke isolate karo.
Result: dy/dx = −(ax+g)/(by+f)
(ii) x + tan(xy) = 0:
tan(xy) mein Chain Rule aur xy mein Product Rule dono use hoti hain sath. Phir dy/dx ko algebra se isolate karte hain.
Result: dy/dx = −[cos²(xy) + y]/x
Q16. Inverse Trigonometric Functions ki Differentiation
(i) sin⁻¹(x/a):
Standard formula d/dx[sin⁻¹(u)] = 1/√(1-u²) · u’ use karte hain, jahan u = x/a.
Result: dy/dx = 1/√(a²−x²)
(ii) csc⁻¹[(x²+1)/(x²−1)]:
Ye thora mushkil hai — formula d/dx[csc⁻¹(u)] = -1/(|u|√(u²-1)) · u’ use hoti hai. Beech mein algebra ke through simplify karna padta hai (x²+1 hamesha positive hota hai, isliye |u| ko simplify karna asaan ho jata hai).
Result: dy/dx = 2x/[|x|(x²+1)], matlab x>1 ke liye +2/(x²+1) aur x<−1 ke liye −2/(x²+1)
(iii) cot⁻¹(√x) + √(cot⁻¹x):
Do alag terms, dono mein inverse trig ki formula aur Chain Rule use hoti hai.
Result: dy/dx = −1/[2√x(1+x)] − 1/[2(1+x²)√(cot⁻¹x)]
(iv) (tan t)⁻¹ + 1/(tan⁻¹t):
Zaroori Note: (tan t)⁻¹ ka matlab hai 1/(tan t) = cot t (reciprocal), na ke inverse tangent function! Ye do alag cheezein hain. Isay samajhna important hai warna ghalti ho sakti hai.
cot t ki derivative -csc²t hoti hai, aur (tan⁻¹t)⁻¹ ki derivative Chain Rule se nikalti hai.
Result: dy/dt = −csc²t − 1/[(1+t²)(tan⁻¹t)²]
Important Tips is exercise se:
- Hamesha check karo pehle ke expression simplify ho sakta hai (jaise Q5).
- Notation ka farq samajhna zaroori hai — jaise (tan t)⁻¹ vs tan⁻¹t (Q16-iv).
- Chain Rule “andar se bahar” lagti hai — jab function ke andar function ho.
- Parametric aur Implicit differentiation mein hamesha formula yaad rakho: dy/dx = (dy/dt)/(dx/dt) ya phir dy/dx ko algebra se isolate karna.