What is the derivative of (x^{-2}+x^4-6/(3x^3+4x^4)) ?

The derivative of (x^{-2}+x^4-6/(3x^3+4x^4)) is (3x^6+96x^3+54x^2-24x-15)/((3x^3+4x^4)^2)

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

The first derivative of (x^{-2}+x^4-6/(3x^3+4x^4)) is (3x^6+96x^3+54x^2-24x-15)/((3x^3+4x^4)^2)

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

\frac{3 x ^{6} + 96 x ^{3} + 54 x ^{2} - 24 x - 15}{( 3 x ^{3} + 4 x ^{4} ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (3 x^{3} + 4 x^{4}) = 9 x^{2} + 16 x^{3}

= \frac{( 4 x ^{3} - \frac{2}{x ^{3}} ) ( 3 x ^{3} + 4 x ^{4} ) - ( 9 x ^{2} + 16 x ^{3} ) ( x ^{- 2} + x ^{4} - 6 )}{( 3 x ^{3} + 4 x ^{4} ) ^{2}}

Simplify

\frac{( 4 x ^{3} - \frac{2}{x ^{3}} ) ( 3 x ^{3} + 4 x ^{4} ) - ( 9 x ^{2} + 16 x ^{3} ) ( x ^{- 2} + x ^{4} - 6 )}{( 3 x ^{3} + 4 x ^{4} ) ^{2}} : \frac{3 x ^{6} + 96 x ^{3} + 54 x ^{2} - 24 x - 15}{( 3 x ^{3} + 4 x ^{4} ) ^{2}}

= \frac{3 x ^{6} + 96 x ^{3} + 54 x ^{2} - 24 x - 15}{( 3 x ^{3} + 4 x ^{4} ) ^{2}}

What is the derivative of (x^{-2}+x^4-6/(3x^3+4x^4)) ?
The derivative of (x^{-2}+x^4-6/(3x^3+4x^4)) is (3x^6+96x^3+54x^2-24x-15)/((3x^3+4x^4)^2)
What is the first derivative of (x^{-2}+x^4-6/(3x^3+4x^4)) ?
The first derivative of (x^{-2}+x^4-6/(3x^3+4x^4)) is (3x^6+96x^3+54x^2-24x-15)/((3x^3+4x^4)^2)