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

The derivative of (x^4+6/(3-4x^{-4)}) is (x^3(12x^8-32x^4-96))/((3x^4-4)^2)

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

The first derivative of (x^4+6/(3-4x^{-4)}) is (x^3(12x^8-32x^4-96))/((3x^4-4)^2)

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

\frac{x ^{3} ( 12 x ^{8} - 32 x ^{4} - 96 )}{( 3 x ^{4} - 4 ) ^{2}}

Solution steps

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

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

\frac{d}{d x} (3 - 4 x^{- 4}) = \frac{16}{x ^{5}}

= \frac{4 x ^{3} ( 3 - 4 x ^{- 4} ) - \frac{16}{x ^{5}} ( x ^{4} + 6 )}{( 3 - 4 x ^{- 4} ) ^{2}}

Simplify

\frac{4 x ^{3} ( 3 - 4 x ^{- 4} ) - \frac{16}{x ^{5}} ( x ^{4} + 6 )}{( 3 - 4 x ^{- 4} ) ^{2}} : \frac{x ^{3} ( 12 x ^{8} - 32 x ^{4} - 96 )}{( 3 x ^{4} - 4 ) ^{2}}

= \frac{x ^{3} ( 12 x ^{8} - 32 x ^{4} - 96 )}{( 3 x ^{4} - 4 ) ^{2}}

\int x^{7} e^{x^{4}} d x

\frac{d}{d x} (cot (2 x - 1))

(x^{2} + y^{2})^{2} = 4 x^{2} y, (1, 1)

\int_{0}^{2} \frac{2 x}{( x ^{2} + 2 ) ^{3}} d x

\int e^{3 x} - 18 e^{- 4 x} d x

What is the derivative of (x^4+6/(3-4x^{-4)}) ?
The derivative of (x^4+6/(3-4x^{-4)}) is (x^3(12x^8-32x^4-96))/((3x^4-4)^2)
What is the first derivative of (x^4+6/(3-4x^{-4)}) ?
The first derivative of (x^4+6/(3-4x^{-4)}) is (x^3(12x^8-32x^4-96))/((3x^4-4)^2)