What is (\partial)/(\partial x)(4/((x+2)^2)) ?

The answer to (\partial)/(\partial x)(4/((x+2)^2)) is -8/((x+2)^3)

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(\partial)/(\partial x)(4/((x+2)^2))

\frac{\partial}{\partial x} (\frac{4}{( x + 2 ) ^{2}})

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

Solution steps

\frac{\partial}{\partial x} (\frac{4}{( x + 2 ) ^{2}})

Take the constant out:

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

= 4 \frac{\partial}{\partial x} (\frac{1}{( x + 2 ) ^{2}})

Apply exponent rule:

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

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

Apply the chain rule:

- \frac{2}{( x + 2 ) ^{3}} \frac{\partial}{\partial x} (x + 2)

= - \frac{2}{( x + 2 ) ^{3}} \frac{\partial}{\partial x} (x + 2)

\frac{\partial}{\partial x} (x + 2) = 1

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

Simplify

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

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

\frac{4}{x}, 3

t \to 1 lim (\frac{t ^{2} - 1}{t - 1})

\int_{1}^{b} \frac{1}{( x + 6 ) ^{2}} d x

f (x) = ln (4 - x^{2})

\int t^{2} sin (βt) d t

What is (\partial)/(\partial x)(4/((x+2)^2)) ?
The answer to (\partial)/(\partial x)(4/((x+2)^2)) is -8/((x+2)^3)