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If [latex]y = \tan^{-1}\left ( \frac{x^{\frac{1}{3}}+a^{\frac{1}{3}}}{1-x^{\frac{1}{3}}a^{\frac{1}{3}}} \right )[/latex], then dy/dx =

(1) 1/3x2/3(1+x2/3) (2) a/3x2/3(1+x2/3) (3) -1/3x2/3(1+x2/3) (4) -a/3x2/3(1+x2/3) Solution: Given y = tan-1 (x1/3 + a1/3)/(1... View Article

[latex]\frac{d}{dx}\left ( x^{3} \tan ^{2}\frac{x}{2}\right )=[/latex]

(1) x3 tan (x/2) sec2(x/2) + 3x tan2(x/2) (2) x3 tan2 (x/2) sec2(x/2) + 3x2 tan2(x/2) (3) x3 tan (x/2) sec2(x/2) + 3x2 tan2(x/2) (4)... View Article

[latex]\frac{d}{dx}\left ( \frac{\sec x+\tan x}{\sec x-\tan x} \right )=[/latex]

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

[latex]\frac{d}{dx}\sqrt{\sec ^{2}x+cosec^{2}\: x}=[/latex]

(1) 4 cosec 2x cot 2x (2) -4 cosec 2x cot 2x (3) 4 cosec x cot 2x (4) none of these Solution: Let y = √( sec2 x + cosec2 x) = √(1/cos2 x +... View Article

[latex]\frac{d}{dx}\sqrt{x\sin x}=[/latex]

(1) (sin x + x cos x )/2√(x sin x) (2) (sin x + x cos x )/√(x sin x) (3) (x sin x + x cos x )/2√(x sin x) (4) (x sin x + x cos x )/√(2x... View Article

[latex]\frac{d}{dx}\left ( x^{2} +\cos x\right )^{4}=[/latex]

(1) 4(x2 + cos x )(2x - sin x) (2) 4(x2 - cos x )3(2x - sin x) (3) 4(x2 + cos x )3(2x - sin x) (4) 4(x2 + cos x )3(2x + sin x)... View Article

If [latex]y = \left ( \frac{1+\tan x}{1-\tan x} \right )[/latex], then dy/dx =

(1) sec2x/(1-tan x)2 (2) 2 sec2x/(1-tan x)2 (3) 2 sec2x/(1+tan x)2 (4) none of these Solution: Given y = (1+tan x)/(1-tan x) Use... View Article

If [latex]y = \sin \left ( \frac{1+x^{2}}{1-x^{2}} \right )[/latex] , then dy/dx =

(1) (4x/(1-x2) ) cos ((1+x2)/(1-x2)) (2) (x/(1-x2) ) cos ((1+x2)/(1-x2)) (3) (4x/(1-x2)2 ) cos ((1+x2)/(1-x2)) (4) (x/(1-x2)2 ) cos... View Article

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

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

If y = (x cot³ x)^3/2, then dy/dx =

(1) (3/2)(x cot3 x)1/2 [ cot3x - 3x cot2x cosec2x ] (2) (3/2)(x cot3 x)1/2 [ cot2x - 3x cot2x cosec2x ] (3) (3/2)(x cot3 x)3/2 [ cot3x - 3x... View Article

If y = [√(a+x) – √(a-x)]/[√(a+x) +√(a-x)], then dy/dx =

(1) ay/x√(a2-x2) (2) ay/√(a2-x2) (3) ay/x√(x2-a2) (4) none of these Solution: Given y = [√(a+x) - √(a-x)]/[√(a+x) +√(a-x)]... View Article

If A = (2^x cot x)/√x, then dA/dx =

(1) (2x/√x) [ log 2 cot x - cosec2 x - (cot x)/2x ] (2) (2x/x) [ log 2 cot x + cosec2 x - (cot x)/2x ] (3) (2x/√x) [ log 2 cot x - cosec2 x... View Article

If y = (tan x + cot x)/(tan x – cot x), then dy/dx =

(1) 2 tan 2x sec 2x (2) tan 2x sec 2x (3) -tan 2x sec 2x (4) -2 tan 2x sec 2x Solution: Given y = (tan x + cot x)/(tan x - cot x)... View Article

If y = t^4/3-3t^-2/3, then dy/dt =

(1) (2t2 + 3)/3t5/3 (2) (2t2 + 3)/t5/3 (3) 2(2t2 + 3)/t5/3 (4) 2(2t2 + 3)/3t5/3 Solution: Given y = t4/3-3t-2/3 dy/dt =... View Article

d/dx sin^-1 (3x – 4x³) =

(1) 3/√(1- x2) (2) -1/√(1- x2) (3) 1 (4) none of these Solution: Let y = sin-1 (3x - 4x3) Put x = sin θ θ = sin-1 x Then y =... View Article

If y = sec^-1 ((√x + 1)/(√x – 1)) + sin^-1 ((√x – 1)/(√x + 1)), then dy/dx =

(1) 0 (2) 1/(√x + 1) (3) 1 (4) none of these Solution: Given y = sec-1((√x + 1)/(√x - 1)) + sin-1 ((√x - 1)/(√x + 1)) = cos-1(√x... View Article

d/dx tan^-1 4√x/(1-4x) =

(1) 1/√x(1+4x) (2) 2/√x(1+4x) (3) 4/√x(1+4x) (4) none of these Solution: Let y = tan-1 4√x/(1-4x) We know tan 2A = 2 tan... View Article

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

(1) -cos [cos (sin x)] sin (cos x) cos x (2) -cos [cos (sin x)] sin (sin x) cos x (3) cos [cos (sin x)] sin (cos x) cos x (4) cos [cos... View Article

If x = φ(t), y = ψ(t), then d²y/dx² is equal to

(1) (φ’ψ’’- ψ’φ’’)/(φ’)2 (2) (φ’ψ’’- ψ’φ’’)/(φ’)3 (3) φ’’/ψ’’ (4) ψ’’/φ’’ Solution: Given x = φ(t) and y = ψ(t). dx/dt = φ’(t)... View Article

Maximum slope of the curve y = – x³ + 3x² + 9x – 27 is

1) 0 2) 12 3) 16 4) 32 Correct option: (2) 12 Solution: Equation of the given curve is: y = - x3 + 3x2 + 9x - 27 Differentiating on both... View Article

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