What is the derivative of f(x)= 1/(3x^2)-5/(2x)+8x-6 ?

The derivative of f(x)= 1/(3x^2)-5/(2x)+8x-6 is 5/(2x^2)-2/(3x^3)+8

What is the first derivative of f(x)= 1/(3x^2)-5/(2x)+8x-6 ?

The first derivative of f(x)= 1/(3x^2)-5/(2x)+8x-6 is 5/(2x^2)-2/(3x^3)+8

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derivative of

f (x) = \frac{1}{3 x ^{2}} - \frac{5}{2 x} + 8 x - 6

\frac{5}{2 x ^{2}} - \frac{2}{3 x ^{3}} + 8

Solution steps

\frac{d}{d x} (\frac{1}{3 x ^{2}} - \frac{5}{2 x} + 8 x - 6)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (\frac{1}{3 x ^{2}}) - \frac{d}{d x} (\frac{5}{2 x}) + \frac{d}{d x} (8 x) - \frac{d}{d x} (6)

\frac{d}{d x} (\frac{1}{3 x ^{2}}) = - \frac{2}{3 x ^{3}}

\frac{d}{d x} (\frac{5}{2 x}) = - \frac{5}{2 x ^{2}}

\frac{d}{d x} (8 x) = 8

\frac{d}{d x} (6) = 0

= - \frac{2}{3 x ^{3}} - (- \frac{5}{2 x ^{2}}) + 8 - 0

Simplify

= \frac{5}{2 x ^{2}} - \frac{2}{3 x ^{3}} + 8

(- 3, \frac{π}{6})

y^{3} + x y^{2} - 5 = x + 3 y^{2}, at x = 3

x = - 7

(- 2, 2)

x^{2} + y^{2} = 9

What is the derivative of f(x)= 1/(3x^2)-5/(2x)+8x-6 ?
The derivative of f(x)= 1/(3x^2)-5/(2x)+8x-6 is 5/(2x^2)-2/(3x^3)+8
What is the first derivative of f(x)= 1/(3x^2)-5/(2x)+8x-6 ?
The first derivative of f(x)= 1/(3x^2)-5/(2x)+8x-6 is 5/(2x^2)-2/(3x^3)+8