FIND DIFFERENTIAL DY FOR EACH OF THE FOLLOWING FUNCTIONS

Problem 1 :

Find differential dy for each of the following functions :

(i)  y  =  (1-2x)³/(3-4x)

Solution :

u  =  (1-2x)³ and v  =  3-4x

u'  =  3(1-2x)² (-2)  ===>  -6(1-2x)²

v'  =  -4

y'  =  [(3-4x)(-6(1-2x)²) - (1-2x)³(-4)]/(3-4x)²

y'  =  [(1-2x)²[-6(3-4x) + 4(1-2x)]/(3-4x)²

y'  =  [(1-2x)²[-18+24x+4-8x]/(3-4x)²

y'  =  [(1-2x)²(16x-14)]/(3-4x)²

y'  =  [2(1-2x)²(8x-7)]/(3-4x)²

(ii)  y  =  (3 + sin2x)^2/3

Solution :

y  =  (3 + sin2x)^2/3

y'  =  (2/3) (3 + sin2x)^{(2/3) - 1} (0+2cos 2x)

y'  =  (4/3) (3 + sin2x)^(-1/3) (cos 2x)

y'  =  (4/3) (cos 2x) [1/(3 + sin2x)^1/3]

(iii)  y  =  e^{x^2-5x+7} cos (x²-1)

Solution :

y  =  e^{x^2-5x+7} cos (x²-1)

u  =  e^{x^2-5x+7} and v  =  cos (x²-1)

u'  =  e^{x^2-5x+7} (2x-5) and v'  =  -sin (x²-1) (2x)

u'  =  (2x-5) e^{x^2-5x+7}  and v'  =  -2xsin (x²-1)

=  e^{x^2-5x+7} (-2xsin (x²-1)) + cos (x²-1)(2x-5) e^{x^2-5x+7}

=  e^{x^2-5x+7}[-2xsin (x²-1) + cos (x²-1)(2x-5)]

=  e^{x^2-5x+7}[cos (x²-1)(2x-5)-2xsin (x²-1)]

Problem 2 :

Find df for f(x)  =  x² + 3x and evaluate it for

(i) x = 2 and dx = 0.1 (ii) x = 3 and dx = 0.02

Solution :

(i)  f(x)  =  x² + 3x

y  =  x² + 3x

Differentiating with respect to x.

dy/dx  =  2x+3

dy  =  (2x+3) dx

Applying x  =  2 and dx  =  0.1

dy  =  (2(2) + 3)(0.1)

dy  =  7(0.1)

dy  =  0.7

(ii) x = 3 and dx = 0.02

dy  =  (2(3) + 3)(0.02)

dy  =  9(0.02)

dy  =  0.18

Problem 3 :

Find ∆f and df for the function f for the indicated value of x and ∆x and compare

(i)  f(x)  =  x³ - 2x² ; x  =  2, ∆x  =  dx  =  0.5

Solution :

f(x)  =  x³ - 2x²

dy/dx  =  3x² - 4x

dy  =  (3x² - 4x) dx

dy  =  (3(2)² - 4(2)) (0.5)

dy  =  (12-8) (0.5)

dy  =  4 (0.5)

dy  =  2

(ii)  f(x)  =  x²+2x+3; x  =  -0.5, ∆x  =  dx  =  0.1

Solution :

f(x)  =  y  =  x²+2x+3

dy/dx  =  2x+2

dy  =  (2x+2) dx

dy  =  (2(0.5)+2) (0.1)

dy  =  (3) (0.1)

dy  =  0.3

Problem 4 :

Assuming log₁₀ e  =  0.4343, find the approximate value of log ₁₀ 1003

Solution :

1003  =  1000 + 3

f(1000 + 3)  =  y + dy

y  =  log x

dy  =  (1/x) dx

=  log 1000 + (1/1000) x 3 x log₁₀ e

=  log₁₀ 10³ + (1/1000) x 3 x log₁₀ e

=  3 (1) + (1/1000) x 3 x 0.4343

=  3 + 0.0013029

=  3.0013029

Problem 5 :

If x³y² = (x - y)⁵, find dy/dx at (1, 2).

Solution :

The given function is implicit function.

x³ (2ydy/dx) + y² (3x²) = 5(x - y)⁴ (1 - dy/dx)

2x³y (dy/dx) + 3x²y²) = 5(x - y)⁴ - 5(x - y)⁴ (dy/dx)

2x³y (dy/dx) + 5(x - y)⁴ (dy/dx) = 5(x - y)⁴  -  3x²y²

(dy/dx)[2x³y + 5(x - y)⁴] = 5(x - y)⁴  - 3x²y²

dy/dx = [5(x - y)⁴  - 3x²y²]/[2x³y + 5(x - y)⁴]

Slope at (1, 2)

= [5(1 - 2)⁴ - 3(1)²2²]/[2(1)³(2) + 5(1 - 2)⁴]

= [5(-1)⁴ - 12]/[2(2) + 5(-1)⁴]

= (5 - 12)/(4 + 5)

= -7/9

Problem 6 :

x = 2t + 5 and y = t² - 5, then dy/dx = ?

Solution :

The given functions are parametric functions.

x = 2t + 5 and y = t² - 5

dy/dx = (dy/dt) / (dx/dt)

= 2t/2

= t

Problem 7 :

The cost function for the production of x units of a commodity is given by

C(x) = 2x³ - 15x² + 36x + 15

The cost will be minimum when x is equal to 

a)  3     b)  2    c)  1    d)  4

Solution :

C(x) = 2x³ - 15x² + 36x + 15

C'(x) = 2(3x²) - 15(2x) + 36(1) + 0

C'(x) = 6x² - 30x + 36

C'(x) = 0

6x² - 30x + 36 = 0

x² - 5x + 6 = 0

(x - 2) (x - 3) = 0

x = 2 and x = 3

C''(x) = 6(2x) - 30(1)

= 12x - 30

When x = 2

C''(2) = 12(2) - 30

= 24 - 30

= -6 < 0 maximum at x = 2

When x = 3

C''(3) = 12(3) - 30

= 36 - 30

= 6 > 0 minimum at x = 3

The cost will be minimum when x = 3.

Problem 8 :

For the functions y = x³ - 3x, the value of d²y/dx² at dy/dx is zero, is

a)  ± 1      b)  ±3      c)  ± 6      d)  none

Solution :

y = x³ - 3x

dy/dx = 3x² - 3(1)

dy/dx = 3x² - 3

dy/dx = 0

3x² - 3 = 0

x² = 1

x = -1 and 1

d²y/dx² = 3(2x) - 0

= 6x

When x = -1

d²y/dx² = 6(-1) ==> -6

When x = 1

d²y/dx² = 6(1) ==> 6

So, the possible values of x are -6 and 6. So, option c is correct.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.