What is the derivative of 3/((4x^2-1^4)) ?

The derivative of 3/((4x^2-1^4)) is -(96x)/((4x^2-1)^5)

What is the first derivative of 3/((4x^2-1^4)) ?

The first derivative of 3/((4x^2-1^4)) is -(96x)/((4x^2-1)^5)

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\frac{d}{d x} (\frac{3}{( 4 x ^{2} - 1 ) ^{4}})

- \frac{96 x}{( 4 x ^{2} - 1 ) ^{5}}

Solution steps

\frac{d}{d x} (\frac{3}{( 4 x ^{2} - 1 ) ^{4}})

Take the constant out:

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

= 3 \frac{d}{d x} (\frac{1}{( 4 x ^{2} - 1 ) ^{4}})

Apply exponent rule:

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

= 3 \frac{d}{d x} ((4 x^{2} - 1)^{- 4})

Apply the chain rule:

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

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

\frac{d}{d x} (4 x^{2} - 1) = 8 x

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

Simplify

3 (- \frac{4}{( 4 x ^{2} - 1 ) ^{5}} \cdot 8 x) : - \frac{96 x}{( 4 x ^{2} - 1 ) ^{5}}

= - \frac{96 x}{( 4 x ^{2} - 1 ) ^{5}}

\frac{d}{d x} (x^{3} + \frac{6}{7} x - 10)

\int 7 tan^{3} (2 x) sec^{5} (2 x) d x

\int_{1}^{2} 5 e^{\frac{1}{x}} d x

\frac{20}{( s ^{2} + 4 ) ( s - 4 )}

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

What is the derivative of 3/((4x^2-1^4)) ?
The derivative of 3/((4x^2-1^4)) is -(96x)/((4x^2-1)^5)
What is the first derivative of 3/((4x^2-1^4)) ?
The first derivative of 3/((4x^2-1^4)) is -(96x)/((4x^2-1)^5)