What is the derivative of (4x^3-7x+8/x) ?

The derivative of (4x^3-7x+8/x) is (8x^3-8)/(x^2)

What is the first derivative of (4x^3-7x+8/x) ?

The first derivative of (4x^3-7x+8/x) is (8x^3-8)/(x^2)

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\frac{d}{d x} (\frac{4 x ^{3} - 7 x + 8}{x})

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

Solution steps

\frac{d}{d x} (\frac{4 x ^{3} - 7 x + 8}{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} - 7 x + 8 ) x - \frac{d x}{d x} ( 4 x ^{3} - 7 x + 8 )}{x ^{2}}

\frac{d}{d x} (4 x^{3} - 7 x + 8) = 12 x^{2} - 7

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

= \frac{( 12 x ^{2} - 7 ) x - 1 \cdot ( 4 x ^{3} - 7 x + 8 )}{x ^{2}}

Simplify

\frac{( 12 x ^{2} - 7 ) x - 1 \cdot ( 4 x ^{3} - 7 x + 8 )}{x ^{2}} : \frac{8 x ^{3} - 8}{x ^{2}}

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

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

\int e^{- s \cdot t} d t

t^{2} y^{^{''}} + 2 t y^{^{'}} - 2 y = 0

\int \frac{4 x + 3}{2 x + 1} d x

y = x^{3} - 16 x + 2

What is the derivative of (4x^3-7x+8/x) ?
The derivative of (4x^3-7x+8/x) is (8x^3-8)/(x^2)
What is the first derivative of (4x^3-7x+8/x) ?
The first derivative of (4x^3-7x+8/x) is (8x^3-8)/(x^2)