What is the derivative of (4x)/((x^3-3)^2) ?

The derivative of (4x)/((x^3-3)^2) is (4(-5x^3-3))/((x^3-3)^3)

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

The first derivative of (4x)/((x^3-3)^2) is (4(-5x^3-3))/((x^3-3)^3)

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derivative of

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

\frac{4 ( - 5 x ^{3} - 3 )}{( x ^{3} - 3 ) ^{3}}

Solution steps

\frac{d}{d x} (\frac{4 x}{( x ^{3} - 3 ) ^{2}})

Take the constant out:

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

= 4 \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}}

= 4 \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)

= 4 \cdot \frac{1 \cdot ( x ^{3} - 3 ) ^{2} - 6 x ^{2} ( x ^{3} - 3 ) x}{( ( x ^{3} - 3 ) ^{2} ) ^{2}}

Simplify

4 \cdot \frac{1 \cdot ( x ^{3} - 3 ) ^{2} - 6 x ^{2} ( x ^{3} - 3 ) x}{( ( x ^{3} - 3 ) ^{2} ) ^{2}} : \frac{4 ( - 5 x ^{3} - 3 )}{( x ^{3} - 3 ) ^{3}}

= \frac{4 ( - 5 x ^{3} - 3 )}{( x ^{3} - 3 ) ^{3}}

x + tan (x)

\int \frac{x ^{2}}{x ^{2} + y ^{2}} d y

\int x \frac{1}{1} d x

\frac{d}{d t} (b sin (wt))

\frac{d}{d x} (w \cdot sin (w c t) \cdot cos (w x))

What is the derivative of (4x)/((x^3-3)^2) ?
The derivative of (4x)/((x^3-3)^2) is (4(-5x^3-3))/((x^3-3)^3)
What is the first derivative of (4x)/((x^3-3)^2) ?
The first derivative of (4x)/((x^3-3)^2) is (4(-5x^3-3))/((x^3-3)^3)