What is the derivative of f(x)=(x^3+2)/(3x) ?

The derivative of f(x)=(x^3+2)/(3x) is (2x^3-2)/(3x^2)

What is the first derivative of f(x)=(x^3+2)/(3x) ?

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

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derivative of

f (x) = \frac{x ^{3} + 2}{3 x}

\frac{2 x ^{3} - 2}{3 x ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{3} + 2}{3 x})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

\frac{d x}{d x} = 1

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

Simplify

\frac{1}{3} \cdot \frac{3 x ^{2} x - 1 \cdot ( x ^{3} + 2 )}{x ^{2}} : \frac{2 x ^{3} - 2}{3 x ^{2}}

= \frac{2 x ^{3} - 2}{3 x ^{2}}

f (x) = \frac{8}{x}, at x = 36

x \to 2 lim (3 x - 7)

(- e^{t}) + \frac{1}{2} (sin (t)) + \frac{1}{2} (cos (t))

x \to - \frac{π}{2} lim (sec (x))

\frac{d}{d x} ((sin (x) + x)^{2} + 7 cos (5 x))

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