What is the derivative of (8x^3-6/(x^2)) ?

The derivative of (8x^3-6/(x^2)) is (4(2x^3+3))/(x^3)

What is the first derivative of (8x^3-6/(x^2)) ?

The first derivative of (8x^3-6/(x^2)) is (4(2x^3+3))/(x^3)

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (8 x^{3} - 6) = 24 x^{2}

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

= \frac{24 x ^{2} x ^{2} - 2 x ( 8 x ^{3} - 6 )}{( x ^{2} ) ^{2}}

Simplify

\frac{24 x ^{2} x ^{2} - 2 x ( 8 x ^{3} - 6 )}{( x ^{2} ) ^{2}} : \frac{4 ( 2 x ^{3} + 3 )}{x ^{3}}

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

y^{^{''}} + 4 y = 3 x - 1

\int 2 x sin^{2} (x) d x

y^{^{'}} = - x

x^{2} \frac{d y}{d x} - 2 x y = 3 y^{4}

y = (\frac{x ^{2} + 6}{x ^{2} - 6})^{3}

What is the derivative of (8x^3-6/(x^2)) ?
The derivative of (8x^3-6/(x^2)) is (4(2x^3+3))/(x^3)
What is the first derivative of (8x^3-6/(x^2)) ?
The first derivative of (8x^3-6/(x^2)) is (4(2x^3+3))/(x^3)