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

The derivative of 2/(2x^4-5) is -(16x^3)/((2x^4-5)^2)

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

The first derivative of 2/(2x^4-5) is -(16x^3)/((2x^4-5)^2)

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d x} (\frac{1}{2 x ^{4} - 5})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= 2 \frac{d}{d x} ((2 x^{4} - 5)^{- 1})

Apply the chain rule:

- \frac{1}{( 2 x ^{4} - 5 ) ^{2}} \frac{d}{d x} (2 x^{4} - 5)

= - \frac{1}{( 2 x ^{4} - 5 ) ^{2}} \frac{d}{d x} (2 x^{4} - 5)

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

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

Simplify

2 (- \frac{1}{( 2 x ^{4} - 5 ) ^{2}} \cdot 8 x^{3}) : - \frac{16 x ^{3}}{( 2 x ^{4} - 5 ) ^{2}}

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

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

\int \frac{2 x + 3}{x ^{2} + 2 x + 5} d x

\frac{\partial}{\partial x} (e^{- \frac{y}{x ^{2} + y ^{2} + 1}})

(- 3.7) (0.3)

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

What is the derivative of 2/(2x^4-5) ?
The derivative of 2/(2x^4-5) is -(16x^3)/((2x^4-5)^2)
What is the first derivative of 2/(2x^4-5) ?
The first derivative of 2/(2x^4-5) is -(16x^3)/((2x^4-5)^2)