What is the derivative of 3/(x^2+9) ?

The derivative of 3/(x^2+9) is -(6x)/((x^2+9)^2)

What is the first derivative of 3/(x^2+9) ?

The first derivative of 3/(x^2+9) is -(6x)/((x^2+9)^2)

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\frac{d}{d x} (\frac{3}{x ^{2} + 9})

- \frac{6 x}{( x ^{2} + 9 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{3}{x ^{2} + 9})

Take the constant out:

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

= 3 \frac{d}{d x} (\frac{1}{x ^{2} + 9})

Apply exponent rule:

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

= 3 \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

= 3 (- \frac{1}{( x ^{2} + 9 ) ^{2}} \cdot 2 x)

Simplify

3 (- \frac{1}{( x ^{2} + 9 ) ^{2}} \cdot 2 x) : - \frac{6 x}{( x ^{2} + 9 ) ^{2}}

= - \frac{6 x}{( x ^{2} + 9 ) ^{2}}

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

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

\frac{\partial}{\partial x} (cos^{6} (x^{8} y^{6}))

y = 2 x - x^{2}, y = 2 x - 4

x \to 9 lim \frac{( x ^{\frac{1}{2}} - 3 ) ( x + 1 )}{x ^{2} - 8 x - 9}

What is the derivative of 3/(x^2+9) ?
The derivative of 3/(x^2+9) is -(6x)/((x^2+9)^2)
What is the first derivative of 3/(x^2+9) ?
The first derivative of 3/(x^2+9) is -(6x)/((x^2+9)^2)