What is the derivative of f(x)=(6x)/((x^3-9)^2) ?

The derivative of f(x)=(6x)/((x^3-9)^2) is (6(-5x^3-9))/((x^3-9)^3)

What is the first derivative of f(x)=(6x)/((x^3-9)^2) ?

The first derivative of f(x)=(6x)/((x^3-9)^2) is (6(-5x^3-9))/((x^3-9)^3)

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derivative of

f (x) = \frac{6 x}{( x ^{3} - 9 ) ^{2}}

\frac{6 ( - 5 x ^{3} - 9 )}{( x ^{3} - 9 ) ^{3}}

Solution steps

\frac{d}{d x} (\frac{6 x}{( x ^{3} - 9 ) ^{2}})

Take the constant out:

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

= 6 \frac{d}{d x} (\frac{x}{( x ^{3} - 9 ) ^{2}})

Apply the Quotient Rule:

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

= 6 \cdot \frac{\frac{d x}{d x} ( x ^{3} - 9 ) ^{2} - \frac{d}{d x} ( ( x ^{3} - 9 ) ^{2} ) x}{( ( x ^{3} - 9 ) ^{2} ) ^{2}}

\frac{d x}{d x} = 1

\frac{d}{d x} ((x^{3} - 9)^{2}) = 6 x^{2} (x^{3} - 9)

= 6 \cdot \frac{1 \cdot ( x ^{3} - 9 ) ^{2} - 6 x ^{2} ( x ^{3} - 9 ) x}{( ( x ^{3} - 9 ) ^{2} ) ^{2}}

Simplify

6 \cdot \frac{1 \cdot ( x ^{3} - 9 ) ^{2} - 6 x ^{2} ( x ^{3} - 9 ) x}{( ( x ^{3} - 9 ) ^{2} ) ^{2}} : \frac{6 ( - 5 x ^{3} - 9 )}{( x ^{3} - 9 ) ^{3}}

= \frac{6 ( - 5 x ^{3} - 9 )}{( x ^{3} - 9 ) ^{3}}

x \to 3 + lim (2 x + 5)

\frac{d}{d t} (2 t + 4)

\frac{d}{d x} (tan^{4} (\frac{1 - x}{1 + x}))

\frac{d y}{d t} = - 0.4 y + 6.2 e^{- 0.73 t}, y (0) = 0

s l o p e (- 2, 10), (4, - 2)

What is the derivative of f(x)=(6x)/((x^3-9)^2) ?
The derivative of f(x)=(6x)/((x^3-9)^2) is (6(-5x^3-9))/((x^3-9)^3)
What is the first derivative of f(x)=(6x)/((x^3-9)^2) ?
The first derivative of f(x)=(6x)/((x^3-9)^2) is (6(-5x^3-9))/((x^3-9)^3)