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All questions of Higher Order Derivatives for Commerce Exam

Differentiate sin22 + 1) with respect to θ2
  • a)
    sin(2θ2 + 1)
  • b)
    cos(2θ2 + 2)
  • c)
    sin(2θ2 + 2)
  • d)
    cos(2θ2 + 1)
Correct answer is option 'C'. Can you explain this answer?

Gunjan Lakhani answered
y = sin22+1)
v = θ2
dy/d(v) = dydθ/dvdθ
dy/dthη = sin2(V+1)
= 2sin(V+1)⋅cos(V+1)dv/dθ
= 2sin(θ2+1)cos(θ2+1)
= sin2(θ2+1).

If 3 sin(xy) + 4 cos (xy) = 5, then   = .....
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Krishna Iyer answered
3sinxy + 4cosxy = 5
⇒ 5(3/5 sinxy + 4/5 cosxy) = 5 
⇒ (3/5 sinxy + 4/5 cosxy) = 1
now (3/5)²+(4/5)² = 1
    so let, 3/5 =   cosA
             ⇒ 4/5 = sinA
So , (3/5 sinxy + 4/5 cosxy) = 1
     ⇒ (cosAsinxy + sinAcosxy) = 1
     ⇒ sin(A+xy) = 1
     ⇒ A + xy = 2πk + π/2 (k is any integer)
     ⇒ sin⁻¹(4/5) + xy = 2πk + π/2
     differenciating both sides with respect to x
   0 + xdy/dx + y = 0
      dy/dx = -y/x


Correct answer is option 'A'. Can you explain this answer?

Aryan Khanna answered
y = tan-1(1-cosx)/sinx
y = tan-1{2sin2(x/2)/(2sin(x/2)cos(x/2)}
y = tan-1{tan x/2}
y = x/2  => dy/dx = 1/2

  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Poonam Reddy answered
y + sin y = 5x
dy/dx + cos ydy/dx = 5
dy/dx = 5/(1+cos y)

  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Suhani Dangarh answered
Put x=tan thita. then you will get. 2 tan inverse x then differentiate

  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Rocky Gupta answered
X^a y^b = (x + y)^(a + b)
taking ln on both sides :-
alnx + b lny = (a + b) ln(x + y)
diff both sides w.r.t x :-
a/x + by'/y = ( (a + b)/(x + y) ) + (((a + b) y'))/(x + y)
or,
a/x - (a +b)/(x + y) = y'[((a + b) / (x + y)) - b/y]
or,
(ax + ay - ax - bx)/x = y' [ (ay + by - bx - by)/y] (cancel (x + y))
or,
y' = dy/dx = ( y (ay - bx) )/(x( ay - bx)) = y/x
therefore we can easily say that the option (D) is the correct answer

  • a)
    2t
  • b)
    1/2a
  • c)
    -t/2a
  • d)
    t/4a
Correct answer is option 'B'. Can you explain this answer?

Om Desai answered
y = at4,     x = 2at2
dy/dt = 4at3        dx/dt = 4at    => dt/dx = 1/4at
Divide dy/dt by dx/dt, we get
dy/dx = t2
d2 y/dx2 = 2t dt/dx……………….(1)
Put the value of dt/dx in eq(1)
d2 y /dx2 = 2t(1/4at)
= 1/2a

If f (x) = [x sin p x] { where [x] denotes greatest integer function}, then f (x) is
  • a)  
    Continuous at x = 0
  • b) 
     Continuous in (-1, 0)
  • c) 
    Differentiable at x = 1
  • d) 
    Differentiable in (-1, 1)
Correct answer is option 'A'. Can you explain this answer?

Asha Choudhury answered
Solution:

To determine the continuity and differentiability of f(x), we need to evaluate left-hand limit, right-hand limit and the derivative of f(x) at every point in the domain of f(x).

Left-hand limit of f(x):
As x approaches 0 from the left, [x] approaches -1 and sin(px) oscillates between -1 and 1. Therefore, f(x) oscillates between -x and x. Thus, the left-hand limit of f(x) as x approaches 0 is:

lim x→0- f(x) = lim x→0- [x sin(px)] = lim x→0- (-x) = 0.

Right-hand limit of f(x):
As x approaches 0 from the right, [x] approaches 0 and sin(px) oscillates between -1 and 1. Therefore, f(x) oscillates between 0 and x. Thus, the right-hand limit of f(x) as x approaches 0 is:

lim x→0+ f(x) = lim x→0+ [x sin(px)] = lim x→0+ (0) = 0.

Since the left-hand limit and right-hand limit are equal, f(x) is continuous at x=0.

Derivative of f(x):
For x≠0, f(x) is piecewise linear with slope sin(px) and jump discontinuity at every integer multiple of 1/p. Therefore, f(x) is not differentiable at these points.

Since f(x) is continuous at x=0 and not differentiable at any other point, the correct answer is option 'A'.

If f(x) = x + cot x, 
  • a)
    -4
  • b)
    2
  • c)
    4
  • d)
    -2
Correct answer is option 'C'. Can you explain this answer?

Aryan Khanna answered
 f(x) = x + cot x
f’(x) = 1 + (-cosec2 x)
f”(x) = 0 - 2cosec x(-cosec x cot x)
= 2 cosec2 x cot x
f”(π/4) = 2 cosec2 (π/4) cot(π/4)
= 2 [(2)^½]2 (1)
= 4

Differentiate sin x3 with respect to x3
  • a)
    sin x3
  • b)
    cos x3
  • c)
    – cos x3
  • d)
    – sin x3
Correct answer is option 'B'. Can you explain this answer?

Nabanita Bajaj answered
Differentiating sin(x^3) with respect to x^3:

To differentiate sin(x^3) with respect to x^3, we can use the chain rule of differentiation. The chain rule states that if we have a composite function, the derivative is equal to the derivative of the outer function multiplied by the derivative of the inner function.

Mathematically, let's represent sin(x^3) as f(g(x)), where g(x) = x^3 and f(x) = sin(x).

- The derivative of g(x) = x^3 with respect to x^3 is simply 1, as x^3 is the inner function.
- The derivative of f(x) = sin(x) with respect to x is cos(x), as it is a standard result from trigonometry.

Now we can apply the chain rule:

df(g(x))/dx = df(g(x))/dg(x) * dg(x)/dx

Substituting the values:

df(g(x))/dg(x) = df(g(x))/dx^3 = cos(x^3)

dg(x)/dx = dg(x)/dx^3 = 1

Therefore,

df(g(x))/dx = cos(x^3) * 1 = cos(x^3)

So, the correct answer is option 'B', which is cos(x^3).

Find the second derivative of excosx​
  • a)
    -2exsinx
  • b)
    -exsinx
  • c)
    ex(sinx + cosx)
  • d)
    -2excosx
Correct answer is option 'A'. Can you explain this answer?

Rounak Nair answered
**Solution:**

To find the second derivative of the given function, we need to differentiate it twice with respect to x.

First, let's find the first derivative of the function:

f(x) = ex * cosx

Using the product rule, the derivative of f(x) is:

f'(x) = (ex * (-sinx)) + (cosx * ex)
= -ex * sinx + ex * cosx
= ex * (cosx - sinx)

Now, let's find the second derivative of the function. Taking the derivative of f'(x):

f''(x) = (ex * (-sinx)) + (ex * (-cosx))
= -ex * sinx - ex * cosx
= -ex * (sinx + cosx)

Therefore, the second derivative of excosx is -2exsinx, which is option A.

  • a)
  • b)
  • c)
  • d)
    none
Correct answer is option 'B'. Can you explain this answer?

Nikita Singh answered
Difference equation are the equations used in discrete time systems and difference equations are similar to the differential equation in continuous systems.

Derivative of cos with respect to x is
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Let x= cos2A then root of 1+cos2A /2 = cosA .So cos^-1( cosA) =A=(cos^-1 x )/2.Then by using derivative we can find out -1/2root 1-x^2

Differentiate   with respect to x. 
  • a)
    -1
  • b)
  • c)
    π/4 − x
  • d)
    1
Correct answer is option 'A'. Can you explain this answer?

Krishna Iyer answered
(cosx − sinx)/(cosx + sinx) = (1 − tanx)/(1 + tanx)
tan(A − B) = (tanA − tanB)/(1 + tanAtanB)
= tan(π/4−x)
putting this value in question.
tan−1 tan(π/4−x)
π/4 − x.
so d(π/4 − x)/dx = -1

The general solution of
is
  • a)
    (c1 + c2x)e3x 
  • b)
    (c1 + c2 In x)x3 
  • c)
    (c1 + c2 x)x
  • d)
    (c1 + c2 In x)ex3
Correct answer is option 'B'. Can you explain this answer?

Varun Kapoor answered
The operator form of given equation is (x2D2 – 5xD + 9)y = 0 ...(*)
Let x = et ⇒ t = log x D′ ≡ d / dt
D ≡ d / dt
We have x2D2 = D′(D′ – 1) 
xD = D′ 
[D′(D′ – 1) – 5D′ + 9]y = 0 
The A.E. is m2 – 6m + 9 = 0 
(m – 3)2 = 0, m = 3, 3 
The C.F. is y = (c
1
 + c2t)e3t 
The solution of (*) is 
y = (c1 + c2 log x) x3.

Differentiate   with respect to x2
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

If you are studying maths, then you should learn the Quotient Rule for Differentiation, and practice how to use it:

  • a)
    π/2
  • b)
  • c)
    1
Correct answer is option 'C'. Can you explain this answer?

x = a(t+sin t) 
⟹dx/dt = a(1+cos t)
And y = a(1−cos t)
⟹dy/dt = a[0−(−sin t)]
=a sin t
Therefore, dy/dx = a sin t/(a(1+cos t))
= 2sin t/2 cos t/2)/(2cos^2 t/2)
=tan(t/2)
At pi/2 = tan(pi/2) = 1

Find ; x = 20 (cos t + t sin t) and y = 20 ( sin t - t cos t)
  • a)
    Sec2t / 20t
  • b)
    Sec3t / 20t3
  • c)
    Sec3t / 20t
  • d)
    Sec3t / 30t
Correct answer is option 'C'. Can you explain this answer?

Hansa Sharma answered
x = 20(cost + tsint)
differentiate x with respect to t,
dx/dt = 20{d(cost)/dt + d(tsint)/dt]
= 20[-sint + {t. d(sint)/dt + sint.dt/dt}]
= 20[ -sint + tcost + sint]
= -20t.cost
hence, dx/dt = -20t.cost -----(1)
 
y = 20(sint - tcost)
differentiate y with respect to t,
dy/dt = 20[d(sint)/dt - d(tcost)/dt ]
= 20[cost - {t.d(cost)/dt + cost.dt/dt}]
= 20[cost +tsint -cost]
= 20t.sint
hence, dy/dt = 20t.sint -------(2)
 
dividing equations (2) by (1),
dy/dx = 20t.sint/20t.cost 
 dy/dx = tant
now again differentiate with respect to x
d²y/dx² = sec²t. dt/dx ------(3)
now from equation (1),
dx/dt = 20t.cost
so, dt/dx =1/at.cost put it in equation (3),
e.g., d²y/dx² = sec²t. 1/20t.cost
d²y/dx² = sec³t/20t

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