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

The derivative of 4/((x^2-1)^4) is -(32x)/((x^2-1)^5)

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

The first derivative of 4/((x^2-1)^4) is -(32x)/((x^2-1)^5)

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derivative of

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

- \frac{32 x}{( x ^{2} - 1 ) ^{5}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

4 (- \frac{4}{( x ^{2} - 1 ) ^{5}} \cdot 2 x) : - \frac{32 x}{( x ^{2} - 1 ) ^{5}}

= - \frac{32 x}{( x ^{2} - 1 ) ^{5}}

\frac{6}{5 x + 2}, at a = - 1

y^{^{'}} + 6 y = 60, y (1) = 90

f (x) = tan (x^{2} + 3 x - 4)

\int \frac{cos ( u )}{u} d u

\frac{\partial}{\partial y} ((x - y)^{10})

What is the derivative of 4/((x^2-1)^4) ?
The derivative of 4/((x^2-1)^4) is -(32x)/((x^2-1)^5)
What is the first derivative of 4/((x^2-1)^4) ?
The first derivative of 4/((x^2-1)^4) is -(32x)/((x^2-1)^5)