(d/dx)(1/x4 sec x) = (1) [4 cos x - x sin x]/x5 (2) -[x sin x + 4 cos x]/x5 (3) [4 cos x + x sin x]/x5 (4) none of these Solution: Let y = 1/x4 sec x... View Article
If y = x + 1/x, then (1) x2(dy/dx) + xy = 0 (2) x2(dy/dx) + xy + 2 = 0 (3) x2(dy/dx) - xy + 2 = 0 (4) none of these Solution: Given y = x + 1/x dy/dx =... View Article
[latex]\frac{d}{dx}\left ( \sqrt{x}+\frac{1}{\sqrt{x}} \right )^{2}=[/latex] (1) 1 - 1/x2 (2) 1 + 1/x2 (3) 1 - 1/2x (4) none of these Solution: Let y = (√x + 1/√x)2 = x + 1/x + 2 dy/dx = 1 - 1/x2... View Article
If u = sin-1 ((x2+y2)/(x+y)) , then x ∂u/∂x + y ∂u/∂y is equal to (1) sin u (2) tan u (3) cos u (4) cot u Solution: u = sin-1 ((x2+y2)/(x+y)) sin u = (x2+y2)/(x+y)) = x2(1+y2/x2)/x(1+y/x) =... View Article
(d/dx)log tan x = (1) 2 sec 2x (2) 2 cosec 2x (3) sec 2x (4) cosec 2x Solution: Let y = log tan x dy/dx = (1/tan x) sec2x = cos x/sin x cos2 x =... View Article
If z = log (tan x + tan y), then sin 2x(∂z/∂x) + sin 2y(∂z/∂y) is equal to (1) 1 (2) 2 (3) 3 (4) 4 Solution: Given z = log (tan x + tan y) Take antilog on both sides ez = (tan x + tan y) ez ∂z/∂x =... View Article
If z = tan (y + ax) – √(y-ax), then zxx-a2zyy is equal to (1) 0 (2) 2 (3) zx+zy (4) zxzy Solution: Given z = tan (y + ax) - √(y-ax) Differentiate partially w.r.t.x zx = sec2(y + ax) a -... View Article
Derivative of sec-1 (1/1-2x2) with respect to sin-1 (3x – 4x3) is (1) 1/4 (2) 3/2 (3) 2/3 (4) 1 Solution: Let u = sec-1(1/1-2x2) Put x = sin θ θ = sin-1x Then u = sec-1(1/1-2 sin2 θ) =... View Article
If [latex]y=\tan^{-1}\left ( \frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}} \right )[/latex] , then dy/dx is equal to (1) x2/√(1-x4) (2) x2/√(1+x4) (3) x/√(1+x4) (4) x/√(1-x4) Solution: Given y = tan-1 (√(1+x2) - √(1-x2) )/(√(1+x2) + √(1-x2))) Put... View Article
If x = cos θ, y = sin 5θ, then (1-x2)d2y/dx2 – xdy/dx is equal to (1) -5y (2) 5y (3) 25y (4) -25y Solution: Given x = cos θ y = sin 5θ dx/dθ = -sin θ dy/dθ = 5 cos 5θ dy/dx = (dy/dθ)/(dx/dθ)... View Article
d2x/dy2 is equal to (1) (d2y/dx2)-1 (2) -(d2y/dx2)-1(dy/dx)-3 (3) -(d2y/dx2)(dy/dx)-3 (4) none of these Solution: dx/dy = 1/(dy/dx) Use... View Article
If y = sin (logex) then x2 d2y/dx2 + xdy/dx is equal to (1) sin (logex) (2) cos (logex) (3) y2 (4) -y Solution: Given y = sin (logex) Differentiate w.r.t.x dy/dx = cos loge x ×(1/x) =... View Article
If x = loge t, t > 0 and y+1 = t2. Then d2y/dx2 is equal to (1) 4e2x (2) (-½)e-4x (3) (-¾)e-5x (4) 4ex Solution: Given x = loge t So ex = t Given y+1 = t2 (ex)2 = y+1 y+1 = e2x... View Article
If [latex]f(x)=1+nx+\frac{n(n-1)}{2!}x^{2}+\frac{n(n-1)(n-2)}{3!}x^{3}+…x^{n}[/latex], then f’’(1) is equal to (1) n(n-1)2n-1 (2) (n-1)2n-1 (3) n(n-1)2n-2 (4) n(n-1)2n Solution: f(x) = 1 + nx + n(n-1)x2/2! + n(n-1)(n-2)x3/3! +....+xn =... View Article
[latex]y=e^{a\sin^{-1}x}[/latex], then [latex](1-x^{2})y_{n+2}-(2n+1)xy_{n+1}[/latex] is equal to 1) – (n2 + a2)yn 2) (n2 - a2)yn 3) (n2 + a2)yn 4) -(n2 - a2)yn Solution: Given Differentiate w.r.t.x Squaring y12 (1-x2) =... View Article
If f (x) = tan–1 x, Then, f ‘ (x) + f ”(x) = 0, where x is equal to 1) 0 2) 1 3) i 4) -i Solution: f (x) = tan–1 x Differentiate w.r.t.x f’(x) = 1/(1+x2) f’’(x) = ((1+x2)×0 - 2x)/(1+x2)2 =... View Article
If y = sec (tan–1 x), then dy/dx at x = 1 is equal to 1) 1/√2 2) 1/2 3) 1 4) √2 Solution: y = sec (tan–1 x) dy/dx = sec (tan–1 x) tan (tan–1 x) ×1/(1+x2) = sec (tan–1 x) x/(1+x2) At... View Article
If x = a cos3 θ and y = a sin3 θ, 1 + (dy/dx)2 is equal to 1) tan θ 2) tan2 θ 3) 1 4) sec2 θ Solution: Given x = a cos3 θ dx/dθ = -3a cos2 θ sin θ y = a sin3 θ dy/dθ = 3a sin2 θ cos θ... View Article
If x = sin t and y = tan t, then dy/dx is equal to (1) cos3 t (2) 1/cos3 t (3) 1/cos2 t (4) 1/sin2 t Solution: Given x = sin t y = tan t dx/dt = cos t dy/dt = sec2 t dy/dx =... View Article
If y = (tan–1 x)2, then (x2 +1)2 y2 + 2x(x2 + 1)y1 is equal to 1) 4 2) 0 3) 2 4) 1 Solution: y = (tan–1 x)2 Differentiate w.r.t.x dy/dx = 2 tan-1x (1/1+x2) dy/dx = (2 tan-1x)/(1+x2)...(i)... View Article
