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If y = sin^-1 √x, then dy/dx =

(1) 2/√x√(1 - x) (2) -2/√x√(1 - x) (3) 1/2√x√(1 - x) (4) 1/√(1 - x) Solution: Given y = sin-1√x Differentiate w.r.t.x dy/dx =... View Article

d/dx [ log (x + 1/x)] =

(1) x + 1/x (2) (1 + 1/x2)/(x + 1/x) (3) (1 - 1/x2)/(x + 1/x) (4) 1 + 1/x Solution: Let y = log (x + 1/x) Differentiate w.r.t.x dy/dx =... View Article

d/dx[tan^-1 (a+x)/(1-ax)] = …; a is constant, a,x∈R⁺; ax<1

(1) a/(1+x2) (2) -a(1+x2) (3) 1/(1+x2) (4) -1/(1+x2) Solution: Let y = tan-1 (a+x)/(1-ax) = tan-1 a + tan-1 x dy/dx + 0 +... View Article

If y = sin^-1x/√(1-x²), then (1-x²)dy/dx is equal to

(1) x+y (2) 1+xy (3) 1-xy (4) xy-2 Solution: y = sin-1x/√(1-x2) y√(1-x2) = sin-1x Differentiate w.r.t.x (dy/dx)√(1-x2) +... View Article

If y = sin^-1 [(5x+12√(1-x²))/13], then dy/dx is equal to

(1) 1/√(1-x2) (2) -1/√(1-x2) (3) 3/√(1-x2) (4) x/√(1-x2) Solution: Given y = sin-1 [(5x+12√(1-x2))/13] Put x = sin θ θ = sin-1 x... View Article

[latex]\frac{d}{dx}e^{x^{3}}[/latex] is equal to

(1) (2) (3) (4) Solution: Given y = Differentiate w.r.t.x Hence option (2) is the answer.... View Article

If y = log_{sin x}tan x, then dy/dx at π/4 is equal to

(1) 4/log 2 (2) -4 log 2 (3) -4 /log 2 (4) none of these Solution: Given y = logsin x tan x We know logb a = log a/log b So y =... View Article

If y = (1+x^1/4) (1+x^1/2)(1-x^1/4), then dy/dx =

(1) 1 (2) -1 (3) x (4) √x Solution: Given y = (1 + x1/4) (1 + x1/2)(1 - x1/4) = (1 - x1/2)(1 + x1/2) = 1 - x dy/dx = -1 Hence... View Article

If y = log log x, then e^y dy/dx =

(1) 1/x (2) 1/x log x (3) 1/log x (4) ey Solution: Given y = log log x dy/dx = (1/log x) (1/x) = 1/x log x ey = elog log x =... View Article

d/dx [ e^ax cos (bx+c)] =

(1) eax (a cos (bx+c) - b sin (bx+c)) (2) eax (a cos (bx+c) + b sin (bx+c)) (3) eax (cos (bx+c) - b sin (bx+c)) (4) none of these... View Article

Let f(x) be a polynomial function of the second degree. If f(1) = f(-1) and a₁, a₂, a₃ are in AP, then f’(a₁), f’(a₂), f’(a₃) are in

(1) AP (2) GP (3) HP (4) none of these Solution: Let f(x) = px2 + qx + r f(1) = p + q + r f(-1) = p - q + r Given f(1) = f(-1) So p + q... View Article

d/dx [log √(sin √e^x)] =

(1) (¼) ex/2 cot ex/2 (2) ex/2 cot ex/2 (3) (¼) ex cot ex (4) (½) ex/2 cot ex/2 Solution: Given y = log √(sin √ex) Differentiate... View Article

[latex]frac{d}{dx}left ( log left ( frac{e^{x}}{1+e^{x}} right ) right )=[/latex]

(1) 1/(1-ex) (2) -1/(1+ex) (3) -1/(1-ex) (4) none of these Solution: Let y = log (ex)/(1+ex) Differentiate w.r.t.x dy/dx =... View Article

d/dx ( log_e x) ( log_a x) =

(1) (1/x) loga x (2) (1/x) logx x (3) (2/x) log x (4) (2/x) loga x Solution: Let y = ( loge x) ( loga x) = (log x/log e) ( log x/ log a)... View Article

If y = sec^-1 1/(1 – 2x²), then dy/dx =

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

If y = tan^-1 ((√a-√x)/(1 + √(ax)), then dy/dx =

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

[latex]\frac{d}{dx}\left ( \cos^{-1}\sqrt{\frac{1+\cos x}{2}} \right )=[/latex]

(1) 1 (2) 1/2 (3) 1/3 (4) none of these Solution: Let y = cos-1 √(1 + cos x)/2 We know 1 + cos x = 2 cos2 (x/2) So y = cos-1 √(2... View Article

If [latex]y = \frac{2(x-\sin x)^{\frac{3}{2}}}{\sqrt{x}}[/latex], then dy/dx =

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

If y = (e^2x+ e^-2x)/(e^2x – e^-2x), then dy/dx =

(1) -8/(e2x - e-2x)2 (2) 8/(e2x - e-2x)2 (3) 4/(e2x - e-2x)2 (4) -4/(e2x - e-2x)2 Solution: Given y = (e2x+ e-2x)/(e2x - e-2x)... View Article

d/dx {e^x log (1 + x²)} =

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

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