What is the derivative of (27/(x^2+9)) ?

The derivative of (27/(x^2+9)) is -(54x)/((x^2+9)^2)

What is the first derivative of (27/(x^2+9)) ?

The first derivative of (27/(x^2+9)) is -(54x)/((x^2+9)^2)

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\frac{d}{d x} (\frac{27}{x ^{2} + 9})

- \frac{54 x}{( x ^{2} + 9 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{27}{x ^{2} + 9})

Take the constant out:

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

= 27 \frac{d}{d x} (\frac{1}{x ^{2} + 9})

Apply exponent rule:

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

= 27 \frac{d}{d x} ((x^{2} + 9)^{- 1})

Apply the chain rule:

- \frac{1}{( x ^{2} + 9 ) ^{2}} \frac{d}{d x} (x^{2} + 9)

= - \frac{1}{( x ^{2} + 9 ) ^{2}} \frac{d}{d x} (x^{2} + 9)

\frac{d}{d x} (x^{2} + 9) = 2 x

= 27 (- \frac{1}{( x ^{2} + 9 ) ^{2}} \cdot 2 x)

Simplify

27 (- \frac{1}{( x ^{2} + 9 ) ^{2}} \cdot 2 x) : - \frac{54 x}{( x ^{2} + 9 ) ^{2}}

= - \frac{54 x}{( x ^{2} + 9 ) ^{2}}

\int_{0}^{7} \frac{2 t}{( t - 8 ) ^{2}} d t

\frac{\partial}{\partial x} (\frac{2 πk t}{ln ( \frac{r}{x} )})

\frac{d}{d t} (t \cdot sin (t))

x y^{2} y^{^{'}} = x + 4

\frac{d}{d x} (3 cot (\frac{x}{8}))

What is the derivative of (27/(x^2+9)) ?
The derivative of (27/(x^2+9)) is -(54x)/((x^2+9)^2)
What is the first derivative of (27/(x^2+9)) ?
The first derivative of (27/(x^2+9)) is -(54x)/((x^2+9)^2)