What is the derivative of (4x^3+2x+5/x) ?

The derivative of (4x^3+2x+5/x) is (8x^3-5)/(x^2)

What is the first derivative of (4x^3+2x+5/x) ?

The first derivative of (4x^3+2x+5/x) is (8x^3-5)/(x^2)

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\frac{d}{d x} (\frac{4 x ^{3} + 2 x + 5}{x})

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

Solution steps

\frac{d}{d x} (\frac{4 x ^{3} + 2 x + 5}{x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 4 x ^{3} + 2 x + 5 ) x - \frac{d x}{d x} ( 4 x ^{3} + 2 x + 5 )}{x ^{2}}

\frac{d}{d x} (4 x^{3} + 2 x + 5) = 12 x^{2} + 2

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

= \frac{( 12 x ^{2} + 2 ) x - 1 \cdot ( 4 x ^{3} + 2 x + 5 )}{x ^{2}}

Simplify

\frac{( 12 x ^{2} + 2 ) x - 1 \cdot ( 4 x ^{3} + 2 x + 5 )}{x ^{2}} : \frac{8 x ^{3} - 5}{x ^{2}}

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

y^{^{'}} sin (y) = x^{4}

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\int \frac{x + 9}{x} d x

\int \frac{5 r ^{2}}{r + 7} d r

\int_{1}^{2} (\frac{1}{x})^{2} d x

What is the derivative of (4x^3+2x+5/x) ?
The derivative of (4x^3+2x+5/x) is (8x^3-5)/(x^2)
What is the first derivative of (4x^3+2x+5/x) ?
The first derivative of (4x^3+2x+5/x) is (8x^3-5)/(x^2)