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

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

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

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

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derivative of

g (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}}

\frac{d}{d x} (sec (x) - x)

y = x^{2} + 4 x - 11, (2, - 1)

f (x) = 1 + x, x = 0

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

x \to \infty lim (x^{2} + x)

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