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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (2 x^{3} - 4 x^{2} + 3) = 6 x^{2} - 8 x

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

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

Simplify

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

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

\int (\frac{1}{( x - y ) ^{\frac{1}{2}}}) d y

\frac{3}{x + 4}

ln (x - 3)

f (t) = 2 e^{- t} cos (2 π t) - 1

\int 2 1 - x^{2} d x

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