differential 16-30

Differentiation

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Question no. 16

If y=f(x) , p= and q= , then what is equal to ?

looks_one -q/p²

looks_two -q/p³

looks_3 1/q

looks_4 q/p²

Option looks_two -q/p³

Solution :
p=
dp/dx = = q
dp = qdx
dx/dy = 1/p
differentiate w.r.t y
d²x/dy² = (-1/p²)(dp/dy)
d²x/dy² = (-1/p²)q(dx/dy) = (-1/p²)q❌(1/-p) = -q/p³

Question no. 17

What is the derivative of log_x5 with respect to log₅x ?

looks_one -(log₅x)^-2

looks_two (log₅x)^-2

looks_3 -(log_x5)^-2

looks_4(log_x5)^-2

option looks_two -(log₅x)^-2

Solution :
log_x5 = log_e5/log_ex
Differentiate w.r.t x

Question no. 18

If y = , then what is equal to ?

looks_one

looks_two

looks_3

looks_4 none of these

option looks_two

Solution :
y =

Question no. 19

If f(x)= log_e[log_ex], then what is f'(e) equal to?

looks_one e^-1

looks_two e

looks_3 1

looks_4 0

option looks_one e^-1

Solution :
f(x)= log_e[log_ex]
f'(x) = (1/logx ) * (1/x )
f'(e) = (1/loge ) * (1/e ) = 1/e = e^-1

Question no. 20

For the curve √x + √y = 1 , what is the value of at (1/4 , 1/4) ?

looks_one 0.5

looks_two 1

looks_3 -1

looks_4 2

option looks_3 -1

Solution :
√x + √y = 1
Differentiate with respect to x

Question no. 21

If y= , what is equal to ?

looks_one

looks_two

looks_3 π/2

looks_4 0

option looks_4 0

Solution :

Question no. 22

If y=a(t + 1/t) and x=a(t - 1/t) , then what is equal to?

looks_one y/x

looks_two x/y

looks_3 -y/x

looks_4 -x/y

option looks_two x/y

Solution :
y=a(t + 1/t)
dy/dt = a(1 - 1/t²) = a(t² - 1)/t²
x = a(t- 1/t)
dx/dt = a(1 + 1/t²) = a(t² + 1)/t²
dy/dx = (t² - 1)/(t² + 1) = x/y

Question no. 23

If y= x^x , then what is equal to?

looks_one x^xlogex

looks_two x^xlogx

looks_3 (1+ logx)

looks_4 x^x (x+ 1/x)

option looks_one x^xlogex

Solution :
y= x^x
log y = xlogx
differentiate with respect to x
(1/y)(dy/dx) = logx + 1
dy/dx = y( logx + 1) = x^x(logx + loge) = x^xlogex

Question no. 24

If y = , then what is equal to?

looks_one x

looks_two xln(10)

looks_3 xlog e

looks_4 None of these

option looks_4 None of these

Solution :
y =
dy/dx = -1/(logx)² ❌ (1/x) = -1/(xlogx²)

Question no. 25

If x= at² and y =2at , then what is equal to?

looks_one

looks_two

looks_3

looks_4

option looks_two

Solution :
x = at²
dx/dt = 2at
y = 2at
dy/dt = 2a
dy/dx = 1/t
= (-1/t²)(dt/dx) = -(-1/t²)❌(1/2at) =

Question no. 26

Let y =t¹⁰ +1 and x = t⁸ + 1 , then is

looks_one 5t/2

looks_two 20t⁸

looks_3 5/16t⁶

looks_4 None of these

option looks_3 5/16t⁶

Solution :
y =t¹⁰ +1
dy/dt = 10t⁹
x = t⁸ + 1
dx/dt = 8t⁷
dy/dx = 10t⁹/8t⁷ = 5t²/4
= (5/4)❌2t ❌dt/dx
= (5t/2)❌(1/8t⁷) = 5/16t⁶

Question no. 27

If x = sintcos2t and y = costsin2t , then at t = π/4 , the value of dy/dx is

looks_one -2

looks_two 2

looks_3 -1/2

looks_4 1/2

option looks_4 1/2

Solution :
x = sintcos2t
dx/dt = costcos2t - 2sintsin2t
y = costsin2t
dy/dt = -sintsin2t - 2sintsin2t
dy/dx = (costcos2t - 2sintsin2t)/(-sintsin2t - 2sintsin2t)
dy/dx at t = π/4 = (cos(π/4)cos(π/2) - 2sin(π/4)sin(π/2))/(-sin(π/4)sin(π/2) - 2sin(π/4)sin(π/2)) = 1/2

Question no. 28

If x=a(t + sint) and y = a(1-cost) , then dy/dx =

looks_one tan(t/2)

looks_two cot(t/2)

looks_3 tan2t

looks_4 tan t

option looks_one tan(t/2)

Solution :
x = a(t + sint)
dx/dt = a(1 + cost)
y = a(1-cost)
dy/dt = asint
dy/dx = a(sint)/(a(1+cost))
dy/dx = sint/(1+cost) = 2sin(t/2)cos(t/2)/2cos²(t/2) = tan(t/2)

Question no. 29

If f(x) = x² -3x , then the points at which f(x) = f'(x) are

looks_one 1,3

looks_two 1 , -3

looks_3 -1, 3

looks_4 None of these

option looks_4 None of these

Solution :
f(x) = x² - 3x
f'(x) = 2x - 3
f(x) = f'(x)
x² - 3x = 2x - 3
x² -5x + 3 = 0
The roots will be irrational.

Question no. 30

If y =a.sinx + bcosx , then y² + (dy/dx)² is a

looks_one Function of x

looks_two Function of y

looks_3 Function of x and y

looks_4 constant

option looks_4 constant

Solution :
y = asinx + bcosx
y' = acosx - bsinx
y² = (asinx + bcosx) = a²sin²x + b²cos²x + 2absinxcosx
(y')² = (acosx - bsinx)² = a²cos²x + b²sin²x - 2absinxcosx
y² + (dy/dx)² = a²sin²x + b²cos²x + 2absinxcosx + a²cos²x + b²sin²x - 2absinxcosx = a² + b²

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