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

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

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

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

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derivative of

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

\frac{2 ( - 5 x ^{3} - 9 )}{( x ^{3} - 9 ) ^{3}}

Solution steps

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

Take the constant out:

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

= 2 \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}}

= 2 \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)

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

Simplify

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

= \frac{2 ( - 5 x ^{3} - 9 )}{( x ^{3} - 9 ) ^{3}}

\frac{d}{d x} (cot (6 x))

f (x) = 7 x + 11, (2, 5)

y^{^{'}} = t y - 2 t - 2

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

y^{^{''}} + 4 y^{^{'}} + 20 y = - 3 sin (2 x)

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