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

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

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

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

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derivative of

f (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}}

\int - \frac{1}{27} sin (3 x) d x

\int \frac{3 - 5 x}{9 - 5 x ^{2}} d x

\frac{d}{d x} (arccos (\frac{6}{x}))

\frac{\partial}{\partial v} (e^{u (v)} + sin (v))

\int_{1}^{2} (- y^{3} + 2 y^{2} + 5 y - 6) d y

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