What is the derivative of f(x)= 1/(x^2-9) ?

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

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

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

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derivative of

f (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}}

\frac{1}{2}

f (x) = e^{9 x s i n (2 x)}

\int x^{2} \cdot e^{- 3 x} d x

\frac{\partial}{\partial x} (- ln (x))

\int \frac{14}{2 x + x x} d x

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