Chapter 1: Further Differentiation — Exercise 2.2
Q1. f'(x) nikalo
(i) f(x) = (x³ + 4x) e^(1/x):
Yahan do functions ka product hai, isliye Product Rule use hogi. Sath hi e^(1/x) ke andar 1/x hai, isliye uski derivative nikalne ke liye Chain Rule bhi lagegi.
- u = x³+4x, jiski derivative u’ = 3x²+4
- v = e^(1/x), jiski derivative Chain Rule se v’ = e^(1/x) · (-1/x²) aati hai (kyunke 1/x ki derivative -1/x² hoti hai)
- Product Rule (u’v + uv’) lagane ke baad common factor e^(1/x) nikal kar simplify karte hain.
Result: f'(x) = e^(1/x) (3x³ − x² + 4x − 4) / x
(ii) f(x) = sin(2^x):
Chain Rule: u = 2^x rakho, phir f(x) = sin u. Yahan zaroori formula yaad rakhna hai: d/dx(aˣ) = aˣ ln a. Isliye u’ = 2^x ln 2.
Phir sin u ki derivative cos u hoti hai, us se u’ multiply karte hain.
Result: f'(x) = 2^x ln 2 · cos(2^x)
(iii) f(x) = e^(x cos x):
Chain Rule: u = x cos x (jo Product Rule se differentiate hota hai: cos x − x sin x). Phir e^u ki derivative khud e^u hoti hai, us se u’ multiply karte hain.
Result: f'(x) = e^(x cos x) (cos x − x sin x)
(iv) f(x) = (eˣ + e⁻ˣ) / (eˣ − e⁻ˣ):
Quotient Rule lagate hain. Numerator expand karne ke baad (a-b)² aur (a+b)² formulas use hote hain, jinme se zyada tar terms cancel ho jate hain aur sirf -4 bachta hai.
Result: f'(x) = −4 / (eˣ − e⁻ˣ)²
Q2. dy/dx nikalo
(i) y = ln(eˣ − e⁻ˣ):
Log function ki derivative ka formula: d/dx[ln u] = u’/u. Yahan u = eˣ − e⁻ˣ, jiski derivative u’ = eˣ + e⁻ˣ hai.
Result: dy/dx = (eˣ + e⁻ˣ)/(eˣ − e⁻ˣ)
(ii) y = log₁₀ √[(1−x²)/(1+x²)]:
Trick: Pehle log properties se simplify karo — square root ko 1/2 power aur division ko subtraction mein badal do:
y = (1/2)log₁₀(1−x²) − (1/2)log₁₀(1+x²)
Phir formula d/dx(log₁₀u) = 1/(u ln10) · u’ use karte hain har term par.
Result: dy/dx = −2x / [(1−x⁴) ln10]
(iii) y = ln(x − √(x²−1)):
u = x − √(x²−1) rakho. Iski derivative nikal kar u’/u formula lagate hain. Beech mein algebra se simplify karne par bohot terms cancel ho jate hain.
Result: dy/dx = −1/√(x² − 1)
(iv) y = ln(1 − cos²x):
Trick: Pehle trigonometric identity use karo: 1 − cos²x = sin²x. Isliye y = ln(sin²x) = 2 ln|sin x| ban jata hai (kyunke power ko log ke aage la sakte hain).
Phir sirf 2 · (cos x/sin x) reh jata hai.
Result: dy/dx = 2 cot x
Sabaq: Log ke andar trigonometric identity dikhe to pehle use simplify karo, phir differentiate karo — kaam bohot asaan ho jata hai.
(v) y = ln x / x²:
Quotient Rule: u = ln x (u’ = 1/x), v = x² (v’ = 2x). Formula (u’v−uv’)/v² lagane se numerator simplify ho kar x(1−2ln x) ban jata hai.
Result: dy/dx = (1 − 2 ln x)/x³
(vi) y = √(x²−1)(2x+1) / (x³−1)^(3/2):
Ye sab se important tareeqa hai is exercise ka: Logarithmic Differentiation.
Jab function bohot saare parts ka product/quotient ho (alag alag powers ke sath), to seedha Product/Quotient Rule lagana bohot mushkil ho jata hai. Is liye trick ye hai:
Step 1: Pehle dono taraf ln lagao:
ln y = (1/2)ln(x²−1) + ln(2x+1) − (3/2)ln(x³−1)
(Yahan log ke rules use hue: multiplication → addition, division → subtraction, power → multiply-outside)
Step 2: Ab dono taraf x ke sath differentiate karo. Left side par y ko x ka function samajh kar Chain Rule lagti hai, jis se y’/y milta hai:
y’/y = x/(x²−1) + 2/(2x+1) − 9x²/[2(x³−1)]
Step 3: Aakhir mein dono taraf y se multiply kar ke y’ ko isolate karo (aur original y ki value wapas dal do):
Result: dy/dx = √(x²−1)(2x+1)/(x³−1)^(3/2) · [x/(x²−1) + 2/(2x+1) − 9x²/(2(x³−1))]
Is Exercise ke Important Tips:
- d/dx(aˣ) = aˣ ln a — ye formula yaad rakho jab base ‘e’ na ho (jaise 2^x).
- Chain Rule har jagah use hoti hai jab function ke andar function ho — chahe wo exponential ho ya logarithmic.
- Log properties (multiplication→addition, power→multiply) hamesha pehle use karo taake calculation asaan ho jaye (jaise Q2 (ii) aur (iv)).
- Logarithmic Differentiation (Q2-vi) sirf tab use karo jab function complicated product/quotient ho jisme kayi factors alag alag powers ke sath hon — ye method bohot time bachata hai.
- Hamesha final answer mein original y ki value wapas substitute karna na bhoolo jab logarithmic differentiation use karo.