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If y sec x + tan x+ x²y = 0, then dy/dx =

(1) (2xy + sec2x + y sec x tan x )/(x2 + sec x) (2) (2xy + sec2x sec x tan x )/(x2 + sec x) (3) -(2xy + sec2x + y sec x tan x )/(x2 + sec... View Article

If tan (x+y) + tan (x-y) = 1, then dy/dx =

(1) [sec2(x+y) + sec2(x-y)]/(sec2 (x+y) - sec2(x-y)) (2) [sec2(x+y) + sec2(x-y)]/(sec2 (x-y) - sec2(x+y)) (3) [sec2(x+y) -... View Article

If x = 2 cos t – cos 2t, y = 2 sin t – sin 2t, then at t = ?/4, dy/dx

(1) (√2+1) (2) √(2-1) (3) √(2+1)/2 (4) none of these Solution: Given x = 2 cos t - cos 2t Differentiate w.r.t.t dx/dt = -2 sin t... View Article

x?(1+y) + y?(1+x) = 0, then dy/dx =

(1) 1+x (2) (1+x)-2 (3) -(1+x)-1 (4) -(1+x)-2 Solution: Given x√(1+y) + y√(1+x) = 0 x√(1+y) = -y√(1+x) Squaring both sides... View Article

If y = (¼) u⁴, u = (?)x³ + 5, then dy/dx

(1) (x2/27)(2x3 + 15)3 (2) (2x/27)(2x3 + 5)3 (3) (2x2/27)(2x3 + 15)3 (4) none of these Solution: Given y = (¼) u4 u = (⅔)x3 + 5... View Article

If cos (x+y) = y sin x, then dy/dx =

(1) -(sin (x+y) + y cos x)/(sin x + sin (x+y)) (2) (sin (x+y) + y cos x)/(sin x + sin (x+y)) (3) (y cos x - sin (x+y))/(sin x - sin (x+y))... View Article

If x = (1 – t²)/(1 + t²) and y = 2t/(1 + t²), then dy/dx =

(1) y/x (2) x/y (3) -x/y (4) -y/x Solution: Given x = (1-t2)/(1+t2) y = 2t/(1+t2) Put t = tan θ θ = tan-1 t x =... View Article

If x = at², y = 2at, then d²y/dx² =

(1) -1/t2 (2) -1/t3 (3) -1/2at3 (4) 1/2at3 Solution: Given x = at2 y = 2at dx/dt = 2at dy/dt = 2a dy/dx = (dy/dt)/(dx/dt )... View Article

If tan y = 2t/(1-t²) and sin x = 2t/(1+t²), then dy/dx =

(1) 2/(1-t2) (2) 1/(1+t2) (3) 1 (4) 2 Solution: Given tan y = 2t/(1 - t2) Put t = tan θ θ = tan-1 t tan y = 2 tan θ/(1 - tan2... View Article

If x = a(t – sin t) and y = a(1 – cos t), then dy/dx =

(1) tan (t/2) (2) -tan (t/2) (3) cot (t/2) (4) -cot (t/2) Solution: Given x = a(t - sin t) dx/dt = a - a cos t = a(1 - cos t) y... View Article

If x = a(cos t + log tan t/2), y = a sin t, then dy/dx =

(1) tan t (2) -tan t (3) cot t (4) -cot t Solution: Given y = a sin t dy/dt = a cos t x = a(cos t + log tan t/2) dx/dt = a... View Article

d/dx [log { e^x ((x-2)/(x+2))^3/4}] equal to

(1) 1 (2) (x2+1)/(x2-4) (3) (x2-1)/(x2-4) (4) ex(x2-1)/(x2-4) Solution: Let y = [log { ex ((x-2)/(x+2))3/4}] = log ex + log... View Article

If y = tan^-1 (√(1 + x²) – 1)/x, then y’(1) =

(1) 1/4 (2) -1/4 (3) 1/2 (4) -1/2 Solution: Given y = tan-1 (√(1 + x2) - 1)/x Put x = tan θ θ = tan-1 x So y = tan-1 (√(1 +... View Article

If 2y = sin^-1 (x + 5y), then dy/dx is equal to

(1) cos 2y - 5 (2) cos 2y + 5 (3) 1/(2 cos 2y - 5) (4) none of these Solution: Given 2y = sin-1 (x + 5y) sin 2y = (x + 5y)... View Article

If y = sec (tan^–1 x), then dy/dx is

1) x/√(1+ x2) 2) -x/√(1+ x2) 3) x/√(1- x2) 4) -x/√(1- x2) Solution: y = sec (tan–1 x) = sec (sec-1√(1+ x2)) = √(1+ x2) Squaring... View Article

d/dx [ tan^-1 (a – x)/(1 + ax) ] =

(1) 1/(1+x2) (2) -1/(1+x2) (3) a/(1+x2) (4) none of these Solution: Let y = [ tan-1 (a - x)/(1 + ax) ] = tan-1 a - tan-1 x... View Article

If y = (a + bx^3/2)/x^5/4 and y’ = 0 at x = 5, then the ratio of a : b is equal to

(1) 5:2 (2) √5 : 1 (3) 1:2 (4) 3:5 Solution: Given y = (a + bx3/2)/x5/4 = ax-5/4 + bx1/4 dy/dx = (-5/4)ax-9/4 + (¼)bx-3/4... View Article

If y = tan^-1 [ sin x + cos x]/[cos x – sin x], then dy/dx is

(1) 0 (2) π/4 (3) 1 (4) 1/2 Solution: Given y = tan-1 [ (sin x + cos x)/(cos x - sin x)] Divide numerator and denominator by cos... View Article

If x = y√(1 – y²), then dy/dx =

If x = y√(1 - y2), then dy/dx = (1) 0 (2) x (3) √(1 - y2)/(1 - 2y2) (4) none of these Solution: Given x = y√(1 - y2) Squaring... View Article

If y = (1 + x²) tan^-1 x – x, then dy/dx =

(1) tan-1 x (2) 2x tan-1 x (3) 2x tan-1 x - 1 (4) 2x /tan-1 x Solution: Given y = (1 + x2) tan-1 x - x Differentiate w.r.t.x... View Article

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