What is the derivative of (x^2+1/(3x^3+4x)) ?

The derivative of (x^2+1/(3x^3+4x)) is (-3x^4-5x^2-4)/((3x^3+4x)^2)

What is the first derivative of (x^2+1/(3x^3+4x)) ?

The first derivative of (x^2+1/(3x^3+4x)) is (-3x^4-5x^2-4)/((3x^3+4x)^2)

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\frac{d}{d x} (\frac{x ^{2} + 1}{3 x ^{3} + 4 x})

\frac{- 3 x ^{4} - 5 x ^{2} - 4}{( 3 x ^{3} + 4 x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} + 1}{3 x ^{3} + 4 x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} + 1 ) ( 3 x ^{3} + 4 x ) - \frac{d}{d x} ( 3 x ^{3} + 4 x ) ( x ^{2} + 1 )}{( 3 x ^{3} + 4 x ) ^{2}}

\frac{d}{d x} (x^{2} + 1) = 2 x

\frac{d}{d x} (3 x^{3} + 4 x) = 9 x^{2} + 4

= \frac{2 x ( 3 x ^{3} + 4 x ) - ( 9 x ^{2} + 4 ) ( x ^{2} + 1 )}{( 3 x ^{3} + 4 x ) ^{2}}

Expand

2 x (3 x^{3} + 4 x) - (9 x^{2} + 4) (x^{2} + 1) : - 3 x^{4} - 5 x^{2} - 4

= \frac{- 3 x ^{4} - 5 x ^{2} - 4}{( 3 x ^{3} + 4 x ) ^{2}}

What is the derivative of (x^2+1/(3x^3+4x)) ?
The derivative of (x^2+1/(3x^3+4x)) is (-3x^4-5x^2-4)/((3x^3+4x)^2)
What is the first derivative of (x^2+1/(3x^3+4x)) ?
The first derivative of (x^2+1/(3x^3+4x)) is (-3x^4-5x^2-4)/((3x^3+4x)^2)