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

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

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

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

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\frac{d}{d x} (\frac{1}{9 x})

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

Solution steps

\frac{d}{d x} (\frac{1}{9 x})

Take the constant out:

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

= \frac{1}{9} \frac{d}{d x} (\frac{1}{x})

Apply exponent rule:

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

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

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

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

Simplify

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

= - \frac{1}{9 x ^{2}}

\int x a^{x} d x

\int \frac{x ^{2}}{( 196 + x ^{2} ) ^{2}} d x

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

n = 1 \sum \infty \frac{2}{4 n ^{2} - 1}

\int \frac{x ^{2} - 1}{x ^{3}} d x

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