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

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

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

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

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

\frac{4 ( x - 3 )}{x ^{4}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{( 3 x ^{2} - 2 ) x ^{3} - 3 x ^{2} ( x ^{3} - 2 x + 4 )}{( x ^{3} ) ^{2}} : \frac{4 ( x - 3 )}{x ^{4}}

= \frac{4 ( x - 3 )}{x ^{4}}

\int x (5 x - 3)^{2} d x

\int e^{3 x} + \frac{1}{x} d x

\frac{d}{d x} (x \frac{d y}{d x})

y^{^{'}} + 3 y = 2 e^{- 3 t}

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

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