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Differential Calculus

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27 Mar 2009 20:05:04 IST

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differentiation

Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Differential Calculus

1) If y = sin^-1√(1+x) + √(1-x)

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then find dy/dx =

a) 1/√1-x^2

b) -1/√1-x^2

c) -1/2√1-x^2

d) none

2) If x=2logcot t and y=tan t + cot t, then (dy/dx)sin 2t +1 =

a) cos^2 t b) sin^2 t (c) cos 2t (d) 2 cos^2 t

3) If sin y = x sin (z+y) and dy/dx=a/x^2+2xb+1 then

a) a=b b) a-b =1 (c) a+b =1 (d) a^2 + b^2 = 1

4) If y = f (2x -1/x^2 +1) and f’(x) = sin x^2. then dy/dx at x=0 equals

a) ½ sin 1 b) sin 1 c) 2 sin 1 d) none

5) If g is inverse of f and f’(x)= 1/1+x^n then g’(x) equals

a) 1 + x^n b) 1+ {f(x)}^n c) 1+{g(x)}^n d) none

6) If y= (1+x)(1+x^2)(1+x^4)………………….(1+x^2n), then at x=0, dy/dx =

a) -1 b) 0 c) 1 d) none

Comments (3)

Srijita

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Joined: 8 Sep 2008

Posts: 66

27 Mar 2009 20:59:58 IST

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ans. to 1st one is (c)

TEJASWI V SHETTY.

Hot goIITian

Joined: 15 Feb 2009

Posts: 186

28 Mar 2009 10:13:08 IST

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d ans to the second one is a). solve it by first differentiating x w.r.t t n then y w.r.t t.divide dy/dt from dx/dt to get the required ans.

hav a nice day.

s faiyaz

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Joined: 30 Mar 2009

Posts: 335

20 Apr 2009 11:13:16 IST

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1) If y = sin^-1√(1+x) + √(1-x)

__________________

then find dy/dx =

a) 1/√1-x^2

b) -1/√1-x^2

c) -1/2√1-x^2

d) none

1. put x = sin 2a then 1 + x = 1 + sin 2a = (sin a + cos a)² or (1 + x )^1/2 = (1 + sin 2a)^1/2 = sina + cos a and similarly (1 - x)^1/2 = sin a - cos a

so y = sin^-1(sin a) = a = 1/2 sin ^-1 x so dy/dx = 1/2(1-x^2)1/2

2) If x=2logcot t and y=tan t + cot t, then (dy/dx)sin 2t +1 =

a) cos^2 t b) sin^2 t (c) cos 2t (d) 2 cos^2 t

dy/dx = (dy/dt)/(dx/dt) = (sec²t-cosec²t)cot t/2(-cosec² t) = cos 2t/sin 2t

(dy/dx)sin 2t + 1 = cos 2t + 1 = 2cos² t

3) If sin y = x sin (z+y) and dy/dx=a/x^2+2xb+1 then

a) a=b b) a-b =1 (c) a+b =1 (d) a^2 + b^2 = 1

sin y = x sin(z+y) = x (sin z cos y + cos z sin y) or 1 = x(sinz cot y + cos z) or 1/x = sin z cot y + cos z or

(1/x – cos z)/sin z = cot y now differentiate w.r.t x both side

4) If y = f (2x -1/x^2 +1) and f’(x) = sin x^2. then dy/dx at x=0 equals

Take 2x – 1/x² + 1 = v now dy/dx = df(v)/dx = df/dv*dv/dx = sin²v*(2+2/x³)

When x = 0, v = 1, dy/dx = sin²1*(2 + 2/0³) so answer in none

a) ½ sin 1 b) sin 1 c) 2 sin 1 d) none

5) If g is inverse of f and f’(x)= 1/1+x^n then g’(x) equals

f(g(x)) = x or df/dx = df/dg*dg/dx = 1 or dg/dx = 1/(df/dg) = 1 + (g(x))ⁿ

a) 1 + x^n b) 1+ {f(x)}^n c) 1+{g(x)}^n d) none

6) If y= (1+x)(1+x^2)(1+x^4)………………….(1+x^2n), then at x=0, dy/dx =

(1+x)(1+x²)(1+x⁴)…..(1+x^p) where p = 2ⁿ, multiplying and dividing by (1-x) we get (1-x^2p)/(1-x)

Differentiate this w.r.t x and putting x = 0, we get ans 1

a) -1 b) 0 c) 1 d) none

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