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

The derivative of (x^3/(x^2-9)) is (x^4-27x^2)/((x^2-9)^2)

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

The first derivative of (x^3/(x^2-9)) is (x^4-27x^2)/((x^2-9)^2)

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

\frac{x ^{4} - 27 x ^{2}}{( x ^{2} - 9 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

\frac{3 x ^{2} ( x ^{2} - 9 ) - 2 x x ^{3}}{( x ^{2} - 9 ) ^{2}} : \frac{x ^{4} - 27 x ^{2}}{( x ^{2} - 9 ) ^{2}}

= \frac{x ^{4} - 27 x ^{2}}{( x ^{2} - 9 ) ^{2}}

\int ln (x) d x

\int \frac{t ^{3} - 8 t + 1}{( 2 t ) ^{4}} d t

y = ln (\frac{3 + e ^{x}}{3 - e ^{x}})

f (x) = (ln (x))^{3}, at x = 9

x \to 0 lim (\frac{2 - 2 cos ( x )}{x})

What is the derivative of (x^3/(x^2-9)) ?
The derivative of (x^3/(x^2-9)) is (x^4-27x^2)/((x^2-9)^2)
What is the first derivative of (x^3/(x^2-9)) ?
The first derivative of (x^3/(x^2-9)) is (x^4-27x^2)/((x^2-9)^2)