What is the derivative of (-6/((x+1)^4)) ?

The derivative of (-6/((x+1)^4)) is (24)/((x+1)^5)

What is the first derivative of (-6/((x+1)^4)) ?

The first derivative of (-6/((x+1)^4)) is (24)/((x+1)^5)

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\frac{d}{d x} (\frac{- 6}{( x + 1 ) ^{4}})

\frac{24}{( x + 1 ) ^{5}}

Solution steps

\frac{d}{d x} (\frac{- 6}{( x + 1 ) ^{4}})

Simplify

\frac{- 6}{( x + 1 ) ^{4}} : - \frac{6}{( x + 1 ) ^{4}}

= \frac{d}{d x} (- \frac{6}{( x + 1 ) ^{4}})

Take the constant out:

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

= - 6 \frac{d}{d x} (\frac{1}{( x + 1 ) ^{4}})

Apply exponent rule:

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

= - 6 \frac{d}{d x} ((x + 1)^{- 4})

Apply the chain rule:

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

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

\frac{d}{d x} (x + 1) = 1

= - 6 (- \frac{4}{( x + 1 ) ^{5}} \cdot 1)

Simplify

- 6 (- \frac{4}{( x + 1 ) ^{5}} \cdot 1) : \frac{24}{( x + 1 ) ^{5}}

= \frac{24}{( x + 1 ) ^{5}}

\frac{\partial}{\partial x} (1 + \frac{c}{x} - \frac{b}{x ^{3}})

k = 1 \sum \infty ln (1 + \frac{1}{k})

x^{2} y^{^{'}} - x y = x^{- 7} y^{\frac{1}{2}}

x \to 2 lim (arctan (\frac{( 2 x ^{2} - 8 )}{3 x ^{2} - 6 x}))

y = - x^{2} - 2 x, y = - 2 x - 4, x = - 3, x = 0

What is the derivative of (-6/((x+1)^4)) ?
The derivative of (-6/((x+1)^4)) is (24)/((x+1)^5)
What is the first derivative of (-6/((x+1)^4)) ?
The first derivative of (-6/((x+1)^4)) is (24)/((x+1)^5)