Answer :
The problems present mathematical series where each equation requires the application of differentiation. The derivative of y = n ln(x) is y' = n/x. The application of derivative rules enables solving these problems.
Here are the derivatives of the given functions:
57. y = -9 ln x
dy/dx = -9/x
58. y = -8 ln x
dy/dx = -8/x
59. y = 7 ln |x|
dy/dx = 7/x (for x > 0)
dy/dx = -7/x (for x < 0)
60. y = 4 ln [x]
dy/dx = 4/x
61. y = [tex]x^{(ln x)}- 3x^2[/tex]
dy/dx = (ln x + 1) * [tex]x^{(ln x)}[/tex] - 6x
62. y = [tex]x^4[/tex] ln x - {[tex]x^3[/tex]}
dy/dx = [tex]4x^3 ln x - 3x^2[/tex]
63. f(x) = ln (9x)
f'(x) = 1/x
64. f(x) = ln (6x)
f'(x) = 1/x
65. f(x) = ln (15x)
f'(x) = 1/x
66. f(x) = ln |10x|
f'(x) = 1/x (for x > 0)
f'(x) = -1/x (for x < 0)
67. g(x) =[tex]x^{(ln (3x))}[/tex]
g'(x) = (ln (3x) + 1) * [tex]x^{(ln (3x))}[/tex] + [tex]x^{(ln (3x)) }[/tex]* (1/x) * 3
68. g(x) = x * ln (7x)
g'(x) = ln (7x) + x * (1/x) * 7
69. g(x) = x + ln (6x)
g'(x) = 1 + 1/x * 6
70. g(x) = [tex]x^0[/tex]* ln (2x)
g'(x) = 0 * ln (2x) + [tex]x^0[/tex] * (1/x) * 2
Please note that ln represents the natural logarithm, [x] represents the floor function, and |x| represents the absolute value of x. Also, remember to apply proper domain restrictions where applicable.
Learn more about the derivative here:
brainly.com/question/35315661
#SPJ11
the question is :
Differentiate.
57. y = -9 ln x
59. y = 7 ln x
58. y=-8 ln x
60. y = 4 ln x
61. y=x Inx-4x+
62. y = x+ Inx - x²
63. f(x) = ln (9x)
64. f(x) = In (6x)
65. f(x) = In 5x
66. f(x) = In 10x
67. g(x) = x5 In (3x)
69. g(x) = x In |6x|
68. g(x) = x² In (7x)
70. g(x) = x In |2x|
