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

The derivative of-4/((x-4^2)) is 8/((x-4)^3)

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

The first derivative of-4/((x-4^2)) is 8/((x-4)^3)

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

\frac{8}{( x - 4 ) ^{3}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

- 4 (- \frac{2}{( x - 4 ) ^{3}} \cdot 1) : \frac{8}{( x - 4 ) ^{3}}

= \frac{8}{( x - 4 ) ^{3}}

\int \frac{2 x + 4}{x ^{2} + 4 x + 13} d x

n = 0 \sum \infty 1400700

n = 1 \sum \infty (\frac{5}{n})^{4}

y = e^{2 x} cos (π x), (0, 1)

f (x) = \frac{x + 7}{8 x}

What is the derivative of-4/((x-4^2)) ?
The derivative of-4/((x-4^2)) is 8/((x-4)^3)
What is the first derivative of-4/((x-4^2)) ?
The first derivative of-4/((x-4^2)) is 8/((x-4)^3)