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

The derivative of (pi^2/(x^2+4x)) is -(pi^2(2x+4))/((x^2+4x)^2)

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

The first derivative of (pi^2/(x^2+4x)) is -(pi^2(2x+4))/((x^2+4x)^2)

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

π^{2} (- \frac{1}{( x ^{2} + 4 x ) ^{2}} (2 x + 4)) : - \frac{π ^{2} ( 2 x + 4 )}{( x ^{2} + 4 x ) ^{2}}

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

3 x y + x^{2} + y^{2}

y^{^{''}} + 3 y^{^{'}} = (4 t + 4) sin (4 t)

(x^{3} + y^{3}) d x - 3 x y^{2} d y = 0

n = 0 \sum \infty \frac{2}{3} n (\frac{1}{3})^{n}

\frac{\partial}{\partial x} (16 - 4 x^{2} - y^{2})

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