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

The derivative of 4/(x^3+3x) is -(12)/(x^4)+3

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

The first derivative of 4/(x^3+3x) is -(12)/(x^4)+3

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\frac{d}{d x} (\frac{4}{x ^{3}} + 3 x)

- \frac{12}{x ^{4}} + 3

Solution steps

\frac{d}{d x} (\frac{4}{x ^{3}} + 3 x)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (\frac{4}{x ^{3}}) + \frac{d}{d x} (3 x)

\frac{d}{d x} (\frac{4}{x ^{3}}) = - \frac{12}{x ^{4}}

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

= - \frac{12}{x ^{4}} + 3

6 cos (6 x)

\int_{- 1}^{0} x^{2} e^{x^{3} + 1} d x

y^{5} \cdot (y^{^{'}} - y) = e^{6 x}

\frac{d y}{d x} = - \frac{2 y ^{2}}{x ^{2}}

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

What is the derivative of 4/(x^3+3x) ?
The derivative of 4/(x^3+3x) is -(12)/(x^4)+3
What is the first derivative of 4/(x^3+3x) ?
The first derivative of 4/(x^3+3x) is -(12)/(x^4)+3