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The differential coefficient of [latex]tan^{-1}sqrt{frac{1-x^{2}}{1+x^{2}}}[/latex] with respect to [latex]cos^{-1}x^{2}[/latex] is

(1) 1/2 (2) -1/2 (3) 1 (4) 0 Solution: Let u = tan-1 √((1 - x2)/(1 + x2)) Put x2 = cos θ So u = tan-1 √((1 - cos θ)/(1 + cos θ))... View Article

The differential coefficient of x⁶with respect to x³ is

(1) 5x2 (2) 3x3 (3) 5x3 (4) 2x3 Solution: Let u = x6 Differentiate w.r.t.x du/dx = 6x5 Let v = x3 dv/dx = 3x2 du/dv =... View Article

The differential coefficient of tan^-1 √x with respect to √x is

(1) 1/(1 + x) (2) 1/2x√(1 + x) (3) 1/2√x(1 + x) (4) none of these Solution: Let u = tan-1 √x Differentiate w.r.t.x du/dx = [1/(1... View Article

The differential coefficient of [latex]tan^{-1}left ( frac{sqrt{1+x^{2}}-1}{x} right )[/latex] with respect to [latex]tan^{-1}x[/latex]

(1) 1/2 (2) -1/2 (3) 1 (4) none of these Solution: Let u = tan-1 (√(1+x2)-1)/x Put x = tan θ u = tan-1 (√(1 + tan2 θ) - 1)/tan θ... View Article

Differential coefficient of [latex]frac{tan^{-1}x}{1+tan^{-1}x}[/latex] w.r.t.[latex]tan^{-1}x[/latex] is

(1) 1/(1 + tan-1 x) (2) -1/(1 + tan-1 x) (3) 1/(1 + tan-1 x)2 (4) -1/2(1 + tan-1 x)2 Solution: Let u = (tan-1x)/(1+tan-1x) v =... View Article

The differential coefficient of [latex]tan^{-1}left ( frac{2x}{1-x^{2}} right )[/latex] w.r.t [latex]sin^{-1}left ( frac{2x}{1+x^{2}} right )[/latex] is

(1) 1 (2) -1 (3) 0 (4) none of these Solution: Let u = tan-1 2x/(1 - x2) Put x = tan θ So u = tan-1 (2 tan θ/(1 - tan2 θ)) =... View Article

Differential coefficient of sin^-1 x w.r.t.cos^-1 √(1 – x²) is

(1) 1 (2) 2 (3) 1/(1 + x2) (4) none of these Solution: Let u = sin-1 x Differentiate w.r.t.x du/dx = 1/√(1 - x2) Let v = cos-1... View Article

Differential coefficient of sin^-1 (1 – x)/(1 + x) w.r.t. √x is

(1) 1/2√x (2) √x/√(1-x) (3) 1 (4) -2/(1+x) Solution: Let u = sin-1(1 - x)/(1 + x) v = √x Differentiate w.r.t.x du/dx = [1/√(1 -... View Article

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

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

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

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

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

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

Differential coefficient of cos^-1 √x with respect to √(1 – x) is

(1) √x (2) -√x (3) 1/√x (4) -1/√x Solution: Let u = cos-1√x v = √(1 - x) Differentiate w.r.t.x du/dx = [-1/√(1 - x)] × 1/2√x =... View Article

If [latex]f(x)=\cot^{-1}\left ( \frac{x^{x}-x^{-x}}{2} \right )[/latex], then f’(1) is equal to

(1) -1 (2) 1 (3) log 2 (4) -log 2 Solution: Given f(x) = cot-1 (xx - x-x)/2 = cot-1 (xx - 1/xx)/2 = cot-1 ((xx)2 - 1)/2xx) =... View Article

If f(x) = x + 2, then f’(f(x)) at x = 4 is

(1) 8 (2) 1 (3) 4 (4) 5 Solution: Given f(x) = x + 2 f(f(x)) = x + 2 + 2 = x + 4 Differentiate w.r.t.x f’(f(x)) = 1 f’(f(x))... View Article

If y² =ax²+ bx + c, then y³ d²y/dx² is

(1) a constant (2) a function of x only (3) a function of y only (4) a function of x and y Solution: Given y2 = ax2 + bx + c... View Article

If [latex]\Delta _{1}=\begin{vmatrix} x & a& b\\ b & x & a\\ a& b & x \end{vmatrix}[/latex] and [latex]\Delta _{2}=\begin{vmatrix} x & b\\ a& x \end{vmatrix}[/latex] are the given determinants, then

(1) ∆1 = 3(∆2)2 (2) d/dx ∆1 = 3(∆2) (3) d/dx ∆1 = 3(∆2)2 (4) ∆1 = 3(∆2)3/2 Solution: Given and ∆1 = x(x2 -ab) - a(xb - a2)... View Article

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