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

The derivative of 1/(1+9x^2) is -(18x)/((1+9x^2)^2)

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

The first derivative of 1/(1+9x^2) is -(18x)/((1+9x^2)^2)

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derivative of

\frac{1}{1 + 9 x ^{2}}

- \frac{18 x}{( 1 + 9 x ^{2} ) ^{2}}

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (1 + 9 x^{2}) = 18 x

= - \frac{1}{( 1 + 9 x ^{2} ) ^{2}} \cdot 18 x

Simplify

- \frac{1}{( 1 + 9 x ^{2} ) ^{2}} \cdot 18 x : - \frac{18 x}{( 1 + 9 x ^{2} ) ^{2}}

= - \frac{18 x}{( 1 + 9 x ^{2} ) ^{2}}

\int \frac{z ^{4} + 10 z ^{3}}{2 z ^{2}} d z

(- 2, - 4), (2, - 4)

\frac{d}{d x} (x \cdot y)

\frac{d}{d x} (arctan (\frac{9}{x}) + arctan (\frac{3}{x}))

f (x) = \frac{3 x}{x ^{2} - 2}, at x = 1

What is the derivative of 1/(1+9x^2) ?
The derivative of 1/(1+9x^2) is -(18x)/((1+9x^2)^2)
What is the first derivative of 1/(1+9x^2) ?
The first derivative of 1/(1+9x^2) is -(18x)/((1+9x^2)^2)