If [latex]f(x)= 3e^{x^{2}}[/latex], then f’(x) – 2xf(x) + ⅓ f(0) – f’(0) = (1) 1 (2) 0 (3) 7/3 (4) none of these Solution: Given f’(x) - 2xf(x) = = = 0 f’(x) - 2xf(x) + ⅓ f(0) - f’(0) = 0 +... View Article
If y = xn log x + x(log x)n, then dy/dx = (1) xn-1( 1 + n log x ) + (log x)n-1 (n + log x) (2) xn-2( 1 + n log x ) + (log x)n-1 (n + log x) (3) xn-1( 1 + n log x ) + (log x)n-1 (n -... View Article
The derivative of tan x – x with respect to x is (1) 1-tan2x (2) tan x (3) -tan2x (4) tan2x Solution: Let y = tan x - x Differentiate w.r.t.x dy/dx = sec2x - 1 = tan2 x Hence... View Article
If y = sin-1√(1-x) + cos-1 √x, then dy/dx = (1) 1/√(x(1-x)) (2) -1/√(x(1-x)) (3) 1/√(x(1+x)) (4) none of these Solution: Given y = sin-1√(1-x) + cos-1 √x dy/dx =... View Article
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 = logsin 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+x1/4) (1+x1/2)(1-x1/4), 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 ey 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 [ eax 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 √ex)] = (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 ( loge x) ( loga 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
