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

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