Higher Order Derivatives

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JEE Main 2011

$\frac{d ^{2} x}{d y ^{2}}$ equals

- (\frac{d ^{2} y}{d x ^{2}})^{- 1} (\frac{d y}{d x})^{- 3}

(B)

(\frac{d ^{2} y}{d x ^{2}})^{- 1} (\frac{d y}{d x})^{- 2}

(C)

- (\frac{d ^{2} y}{d x ^{2}}) (\frac{d y}{d x})^{- 3}

(D)

(\frac{d ^{2} y}{d x ^{2}})^{- 1}

JEE Main 2002

LEVELJEE Main

If $y = (x + 1 + x^{2})^{n}$ , then $(1 + x^{2}) \frac{d ^{2} y}{d x ^{2}} + x \frac{d y}{d x}$ is

(A)

n^{2} y

(B)

- n^{2} y

(C)

- y

(D)

2 x^{2} y

JEE Advanced 2008

LEVELJEE Main

Let $g (x) = lo g (f (x))$ where $f (x)$ is twice differentiable positive function on $(0, \infty)$ such that $f (x + 1) = x f (x)$ . Then, for $N = 1, 2, 3, \dots\dots\dots$ $g^{''} (N + \frac{1}{2}) - g^{''} (\frac{1}{2}) =$

(A)

- 4 {1 + \frac{1}{9} + \frac{1}{25} + \dots + \frac{1}{( 2 N - 1 ) ^{2}}}

(B)

4 {1 + \frac{1}{9} + \frac{1}{25} + \dots + \frac{1}{( 2 N - 1 ) ^{2}}}

(C)

- 4 {1 + \frac{1}{9} + \frac{1}{25} + \dots + \frac{1}{( 2 N + 1 ) ^{2}}}

(D)

4 {1 + \frac{1}{9} + \frac{1}{25} + \dots + \frac{1}{( 2 N + 1 ) ^{2}}}

JEE Advanced 2008

LEVELJEE Main

Let $f$ and $g$ be real valued functions defined on interval $(- 1, 1)$ such that $g^{''} (x)$ is continuous, $g (0) \neq = 0, g^{'} (0) = 0, g^{''} (0) \neq = 0$ , and $f (x) = g (x) sin x$ . \\ STATEMENT-1: $lim_{x \to 0} [g (x) cot x - g (0) csc x] = f^{''} (0)$ and \\ STATEMENT-2: $f^{'} (0) = g (0)$ .

(A)

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

(B)

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

(C)

Statement-1 is True, Statement-2 is False

(D)

Statement-1 is False, Statement-2 is True

JEE Advanced 2007

LEVELJEE Main

$\frac{d ^{2} x}{d y ^{2}}$ equals

(A)

(\frac{d ^{2} y}{d x ^{2}})^{- 1}

(B)

- (\frac{d ^{2} y}{d x ^{2}})^{- 1} (\frac{d y}{d x})^{- 3}

(C)

(\frac{d ^{2} y}{d x ^{2}}) (\frac{d y}{d x})^{- 2}

(D)

- (\frac{d ^{2} y}{d x ^{2}}) (\frac{d y}{d x})^{- 3}

JEE Advanced 2006

LEVELJEE Main

If $F (x) = (f (x /2))^{2} + (g (x /2))^{2}$ where $f^{''} (x) = - f (x)$ and $g (x) = f^{'} (x)$ and given that $F (5) = 5$ , then $F (10)$ is equal to

(A)

5

(B)

10

(C)

0

(D)

15

JEE Advanced 2005

LEVELJEE Main

If $y = y (x)$ and it follows the relation $x cos y + y cos x = π$ then $y^{''} (0) =$

π - 1

If $x^{2} + y^{2} = 1$ then

(A)

y y^{''} - 2 (y^{'})^{2} + 1 = 0

(B)

y y^{''} + (y^{'})^{2} + 1 = 0

(C)

y y^{''} + (y^{'})^{2} - 1 = 0

(D)

y y^{''} + 2 (y^{'})^{2} + 1 = 0

JEE Advanced 2000

LEVELJEE Main

If $x^{2} + y^{2} = 1$ , then

(A)

yy'' - 2(y')^2 + 1 = 0

(B)

yy'' + (y')^2 + 1 = 0

(C)

yy'' + (y')^2 - 1 = 0

(D)

yy'' + 2(y')^2 + 1 = 0

JEE Advanced 1988

LEVELJEE Main

If $y^{2} = P (x)$ , a polynomial of degree 3, then $2 \frac{d}{d x} (y^{3} \frac{d ^{2} y}{d x ^{2}})$ equals

(A)

P^{'''} (x) + P (x)

(B)

P^{'} (x) P^{''} (x)

(C)

P (x) P^{'''} (x)

There exist a function $f (x)$ , satisfying $f (0) = 1, f^{'} (0) = - 1, f (x) > 0$ for all $x$ , and

(A)

f^{''} (x) > 0

for all

x

(B)

- 1 < f^{''} (x) < 0

for all

x

(C)

- 2 \leq f^{''} (x) \leq - 1

for all

x

(D)

f^{''} (x) < - 2

for all

x

JEE Advanced 1982

LEVELJEE Main

Let $f$ be a twice differentiable function such that $f^{''} (x) = - f (x)$ , and $f^{'} (x) = g (x), h (x) = [f (x)]^{2} + [g (x)]^{2}$ . Find $h (10)$ if $h (5) = 11$ .

JEE Advanced 1992

LEVELJEE Main

Let $f (x) = x ∣ x ∣$ . The set of points where $f (x)$ is twice differentiable is .........

JEE Advanced 1983

LEVELJEE Main

If $(a + b x) e^{y / x} = x$ , then prove that $x^{3} \frac{d ^{2} y}{d x ^{2}} = (x \frac{d y}{d x} - y)^{2}$ .

JEE Main 2018 (15 April Shift 1)

LEVELJEE Main

If $x^{2} + y^{2} + sin y = 4$ , then the value of $\frac{d ^{2} y}{d x ^{2}}$ at the point (-2,0) is

JEE Main 2019 (10 January Shift 1)

LEVELJEE Main

Let $f : R \to R$ be a function such that $f (x) = x^{3} + x^{2} f^{'} (1) + x f^{''} (2) + f^{'''} (3), x \in R$ . Then $f (2)$ equal :

JEE Main 2019 (12 April Shift 1)

LEVELJEE Main

If $e^{y} + x y = e$ , the ordered pair $(\frac{d y}{d x}, \frac{d ^{2} y}{d x ^{2}})$ at $x = 0$ is equal to :

(A)

(1/ e, 1/ e^{2})

(B)

(1/ e, - 1/ e^{2})

(C)

(- 1/ e, 1/ e^{2})

(D)

(- 1/ e, - 1/ e^{2})

JEE Main 2020 (9 January Shift 2)

LEVELJEE Main

If $x = 2 sin θ - sin 2 θ$ and $y = 2 cos θ - cos 2 θ, θ \in [0, 2 π]$ , then $\frac{d ^{2} y}{d x ^{2}}$ at $θ = π$ is :

(A)

- \frac{3}{8}

(B)

\frac{3}{4}

(C)

\frac{3}{2}

(D)

- \frac{3}{4}

JEE Main 2020 - 9 Jan (Evening)

LEVELJEE Advanced

If $x = 2 sin θ - sin 2 θ$ and $y = 2 cos θ - cos 2 θ$ , $θ \in [0, 2 π]$ , then $\frac{d ^{2} y}{d x ^{2}}$ at $θ = π$ is:

(A)

\frac{3}{8}

(B)

\frac{3}{2}

(C)

\frac{3}{4}

(D)

- \frac{3}{4}

JEE Main 2020 - 3 Sep (Morning)

LEVELJEE Main

If $y^{2} + lo g_{e} (cos^{2} x) = y, x \in (- \frac{π}{2}, \frac{π}{2})$ , then :

(A)

∣ y^{'} (0) ∣ + ∣ y^{''} (0) ∣ = 3

(B)

∣ y^{'} (0) ∣ + ∣ y^{''} (0) ∣ = 1

(C)

y^{''} (0) = 0

(D)

∣ y^{''} (0) ∣ = 2

JEE Main 2021 (27 August Shift 1)

LEVELJEE Main

If $y^{1/4} + y^{- 1/4} = 2 x$ , and $(x^{2} - 1) \frac{d ^{2} y}{d x ^{2}} + α x \frac{d y}{d x} + β y = 0$ then $∣ α - β ∣$ is equal to .

JEE Main 2021 (26 August Shift 1)

LEVELJEE Main

If $y = y (x)$ is an implicit function of $x$ such that $lo g_{e} (x + y) = 4 x y$ , then $\frac{d ^{2} y}{d x ^{2}}$ at $x = 0$ is equal to .

JEE Main 2021 (16 March Shift 2)

LEVELJEE Main

Let $f : S \to S$ where $S = (0, \infty)$ be a twice differentiable function such that $f (x + 1) = x f (x)$ . If $g : S \to R$ be defined as $g (x) = lo g_{e} f (x)$ , then the value of $∣ g^{''} (5) - g^{''} (1) ∣$ is equal to :

(A)

\frac{205}{144}

(B)

\frac{197}{144}

(C)

\frac{187}{144}

(D)

JEE Main 2022 (25 June Shift 1)

LEVELJEE Main

Let $f (x)$ be a polynomial function such that $f (x) + f^{'} (x) + f^{''} (x) = x^{5} + 64$ . Then, the value of $lim_{x \to 1} \frac{f ( x )}{x - 1}$ is

JEE Main 2022 (24 June Shift 2)

LEVELJEE Main

If $y = tan^{- 1} (sec x^{3} - tan x^{3})$ , $\frac{π}{2} < x^{3} < \frac{3 π}{2}$ , then

(A)

x y^{''} + 2 y^{'} = 0

(B)

x^{2} y^{''} - 6 y + \frac{3 π}{2} = 0

(C)

x^{2} y^{''} - 6 y + 3 π = 0

(D)

x y^{''} - 4 y^{'} = 0

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If F(x)=(f(x/2))2+(g(x/2))2 where f′′(x)=−f(x) and g(x)=f′(x) and given that F(5)=5, then F(10) is equal to

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y2=P(x), a polynomial of degree 3, then 2dxd​(y3dx2d2y​) equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } There exist a function f(x), satisfying f(0)=1,f′(0)=−1,f(x)>0 for all x, and

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f be a twice differentiable function such that f′′(x)=−f(x), and f′(x)=g(x),h(x)=[f(x)]2+[g(x)]2. Find h(10) if h(5)=11.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=x∣x∣. The set of points where f(x) is twice differentiable is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (a+bx)ey/x=x, then prove that x3dx2d2y​=(xdxdy​−y)2.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x2+y2+siny=4, then the value of dx2d2y​ at the point (-2,0) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a function such that f(x)=x3+x2f′(1)+xf′′(2)+f′′′(3),x∈R. Then f(2) equal :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ey+xy=e, the ordered pair (dxdy​,dx2d2y​) at x=0 is equal to :

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x=2sinθ−sin2θ and y=2cosθ−cos2θ, θ∈[0,2π], then dx2d2y​ at θ=π is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y2+loge​(cos2x)=y,x∈(−2π​,2π​), then :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y1/4+y−1/4=2x, and (x2−1)dx2d2y​+αxdxdy​+βy=0 then ∣α−β∣ is equal to .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x) is an implicit function of x such that loge​(x+y)=4xy, then dx2d2y​ at x=0 is equal to .

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=tan−1(secx3−tanx3), 2π​<x3<23π​, then

$\frac{d ^{2} x}{d y ^{2}}$ equals

If $y = (x + 1 + x^{2})^{n}$ , then $(1 + x^{2}) \frac{d ^{2} y}{d x ^{2}} + x \frac{d y}{d x}$ is

Let $g (x) = lo g (f (x))$ where $f (x)$ is twice differentiable positive function on $(0, \infty)$ such that $f (x + 1) = x f (x)$ . Then, for $N = 1, 2, 3, \dots\dots\dots$ $g^{''} (N + \frac{1}{2}) - g^{''} (\frac{1}{2}) =$

$\frac{d ^{2} x}{d y ^{2}}$ equals

If $F (x) = (f (x /2))^{2} + (g (x /2))^{2}$ where $f^{''} (x) = - f (x)$ and $g (x) = f^{'} (x)$ and given that $F (5) = 5$ , then $F (10)$ is equal to

If $y = y (x)$ and it follows the relation $x cos y + y cos x = π$ then $y^{''} (0) =$

If $x^{2} + y^{2} = 1$ then

If $x^{2} + y^{2} = 1$ , then

If $y^{2} = P (x)$ , a polynomial of degree 3, then $2 \frac{d}{d x} (y^{3} \frac{d ^{2} y}{d x ^{2}})$ equals

There exist a function $f (x)$ , satisfying $f (0) = 1, f^{'} (0) = - 1, f (x) > 0$ for all $x$ , and

Let $f$ be a twice differentiable function such that $f^{''} (x) = - f (x)$ , and $f^{'} (x) = g (x), h (x) = [f (x)]^{2} + [g (x)]^{2}$ . Find $h (10)$ if $h (5) = 11$ .

Let $f (x) = x ∣ x ∣$ . The set of points where $f (x)$ is twice differentiable is .........

If $(a + b x) e^{y / x} = x$ , then prove that $x^{3} \frac{d ^{2} y}{d x ^{2}} = (x \frac{d y}{d x} - y)^{2}$ .

If $x^{2} + y^{2} + sin y = 4$ , then the value of $\frac{d ^{2} y}{d x ^{2}}$ at the point (-2,0) is

Let $f : R \to R$ be a function such that $f (x) = x^{3} + x^{2} f^{'} (1) + x f^{''} (2) + f^{'''} (3), x \in R$ . Then $f (2)$ equal :

If $e^{y} + x y = e$ , the ordered pair $(\frac{d y}{d x}, \frac{d ^{2} y}{d x ^{2}})$ at $x = 0$ is equal to :

If $x = 2 sin θ - sin 2 θ$ and $y = 2 cos θ - cos 2 θ, θ \in [0, 2 π]$ , then $\frac{d ^{2} y}{d x ^{2}}$ at $θ = π$ is :

If $x = 2 sin θ - sin 2 θ$ and $y = 2 cos θ - cos 2 θ$ , $θ \in [0, 2 π]$ , then $\frac{d ^{2} y}{d x ^{2}}$ at $θ = π$ is:

If $y^{2} + lo g_{e} (cos^{2} x) = y, x \in (- \frac{π}{2}, \frac{π}{2})$ , then :

If $y^{1/4} + y^{- 1/4} = 2 x$ , and $(x^{2} - 1) \frac{d ^{2} y}{d x ^{2}} + α x \frac{d y}{d x} + β y = 0$ then $∣ α - β ∣$ is equal to .

If $y = y (x)$ is an implicit function of $x$ such that $lo g_{e} (x + y) = 4 x y$ , then $\frac{d ^{2} y}{d x ^{2}}$ at $x = 0$ is equal to .

Let $f (x)$ be a polynomial function such that $f (x) + f^{'} (x) + f^{''} (x) = x^{5} + 64$ . Then, the value of $lim_{x \to 1} \frac{f ( x )}{x - 1}$ is

If $y = tan^{- 1} (sec x^{3} - tan x^{3})$ , $\frac{π}{2} < x^{3} < \frac{3 π}{2}$ , then