What is the derivative of (2x^4+5/(4-4x)) ?

The derivative of (2x^4+5/(4-4x)) is -(6x^4-8x^3-5)/(4(x-1)^2)

What is the first derivative of (2x^4+5/(4-4x)) ?

The first derivative of (2x^4+5/(4-4x)) is -(6x^4-8x^3-5)/(4(x-1)^2)

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\frac{d}{d x} (\frac{2 x ^{4} + 5}{4 - 4 x})

- \frac{6 x ^{4} - 8 x ^{3} - 5}{4 ( x - 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{2 x ^{4} + 5}{4 - 4 x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 2 x ^{4} + 5 ) ( 4 - 4 x ) - \frac{d}{d x} ( 4 - 4 x ) ( 2 x ^{4} + 5 )}{( 4 - 4 x ) ^{2}}

\frac{d}{d x} (2 x^{4} + 5) = 8 x^{3}

\frac{d}{d x} (4 - 4 x) = - 4

= \frac{8 x ^{3} ( 4 - 4 x ) - ( - 4 ) ( 2 x ^{4} + 5 )}{( 4 - 4 x ) ^{2}}

Simplify

\frac{8 x ^{3} ( 4 - 4 x ) - ( - 4 ) ( 2 x ^{4} + 5 )}{( 4 - 4 x ) ^{2}} : - \frac{6 x ^{4} - 8 x ^{3} - 5}{4 ( x - 1 ) ^{2}}

= - \frac{6 x ^{4} - 8 x ^{3} - 5}{4 ( x - 1 ) ^{2}}

\frac{\partial}{\partial x} (25 - x^{2})

\frac{d y}{d x} + \frac{1}{x + 8} y = x^{- 2}

\frac{x ^{2} - 2 x}{( x - 1 ) ^{2}}

f (x) = x^{4} + 9 e^{x}, (0, 9)

e^{x} y d y = (e^{- y} + e^{- 2 x - y}) d x

What is the derivative of (2x^4+5/(4-4x)) ?
The derivative of (2x^4+5/(4-4x)) is -(6x^4-8x^3-5)/(4(x-1)^2)
What is the first derivative of (2x^4+5/(4-4x)) ?
The first derivative of (2x^4+5/(4-4x)) is -(6x^4-8x^3-5)/(4(x-1)^2)