What is the derivative of 8/((1-x)^2) ?

The derivative of 8/((1-x)^2) is (16)/((1-x)^3)

What is the first derivative of 8/((1-x)^2) ?

The first derivative of 8/((1-x)^2) is (16)/((1-x)^3)

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derivative of

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

\frac{16}{( 1 - x ) ^{3}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

8 (- \frac{2}{( 1 - x ) ^{3}} (- 1)) : \frac{16}{( 1 - x ) ^{3}}

= \frac{16}{( 1 - x ) ^{3}}

\int 1 d x

y^{^{''}} + y = 6 e^{- t}, y^{^{'}} (0) = 0, y (0) = 0

y = ∣ 6 x ∣, y = x^{2} - 7

n = 1 \sum \infty \frac{n}{n}

- y

What is the derivative of 8/((1-x)^2) ?
The derivative of 8/((1-x)^2) is (16)/((1-x)^3)
What is the first derivative of 8/((1-x)^2) ?
The first derivative of 8/((1-x)^2) is (16)/((1-x)^3)