What is the derivative of 1/(x^2+9) ?

The derivative of 1/(x^2+9) is -(2x)/((x^2+9)^2)

What is the first derivative of 1/(x^2+9) ?

The first derivative of 1/(x^2+9) is -(2x)/((x^2+9)^2)

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

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

Solution steps

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

Apply exponent rule:

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

= \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

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

Simplify

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

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

(3, - 6), (5, - 5)

y = x^{3} - 1, y = x - 1

n = 0 \sum \infty \frac{3 ^{n}}{5 ^{2 n + 1}}

x \to 1 lim (x + 3 e x - 1 + 1)

1. 8^{2 x}

What is the derivative of 1/(x^2+9) ?
The derivative of 1/(x^2+9) is -(2x)/((x^2+9)^2)
What is the first derivative of 1/(x^2+9) ?
The first derivative of 1/(x^2+9) is -(2x)/((x^2+9)^2)