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

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

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

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

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derivative of

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

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

Solution steps

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

Take the constant out:

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

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

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

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

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

Simplify

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

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

\int \frac{e ^{t a n (x)}}{cos ^{2} ( x )} d x

\int 5 e^{3 x + e^{3 x}} d x

f (x) = sec^{3} (x)

\int (t^{2} + 2) d t

\int \frac{e ^{x}}{e ^{x} ( e ^{x} - 3 )} d x

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