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

The derivative of f(x)= 2/((1-x)^3) is 6/((1-x)^4)

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

The first derivative of f(x)= 2/((1-x)^3) is 6/((1-x)^4)

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derivative of

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial x} (e^{\frac{x ^{2}}{2}})

f (x) = \frac{- 8 x}{x ^{2} + 1}, (1, - 4)

y^{^{''}} + 3.26 y^{^{'}} + 2.66 y = 0

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

\frac{\partial}{\partial y} (x^{2} - 6 y)

What is the derivative of f(x)= 2/((1-x)^3) ?
The derivative of f(x)= 2/((1-x)^3) is 6/((1-x)^4)
What is the first derivative of f(x)= 2/((1-x)^3) ?
The first derivative of f(x)= 2/((1-x)^3) is 6/((1-x)^4)