What is the derivative of f(x)=(22)/((1-x)^3) ?

The derivative of f(x)=(22)/((1-x)^3) is (66)/((1-x)^4)

What is the first derivative of f(x)=(22)/((1-x)^3) ?

The first derivative of f(x)=(22)/((1-x)^3) is (66)/((1-x)^4)

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derivative of

f (x) = \frac{22}{( 1 - x ) ^{3}}

\frac{66}{( 1 - x ) ^{4}}

Solution steps

\frac{d}{d x} (\frac{22}{( 1 - x ) ^{3}})

Take the constant out:

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

= 22 \frac{d}{d x} (\frac{1}{( 1 - x ) ^{3}})

Apply exponent rule:

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

= 22 \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

= 22 (- \frac{3}{( 1 - x ) ^{4}} (- 1))

Simplify

22 (- \frac{3}{( 1 - x ) ^{4}} (- 1)) : \frac{66}{( 1 - x ) ^{4}}

= \frac{66}{( 1 - x ) ^{4}}

y = x^{2} - 2 x, (3, 3)

\frac{d y}{d x} = \frac{x ^{2}}{9} + \frac{7}{9}

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

y^{^{''}} = 8 (y^{^{'}})^{2}

y = e^{x}, y = e^{- 2 x}, x = ln (3)

What is the derivative of f(x)=(22)/((1-x)^3) ?
The derivative of f(x)=(22)/((1-x)^3) is (66)/((1-x)^4)
What is the first derivative of f(x)=(22)/((1-x)^3) ?
The first derivative of f(x)=(22)/((1-x)^3) is (66)/((1-x)^4)