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

The derivative of ((3x^2+x))/((3x^2-x)) is -6/((3x-1)^2)

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

The first derivative of ((3x^2+x))/((3x^2-x)) is -6/((3x-1)^2)

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (3 x^{2} + x) = 6 x + 1

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

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

Simplify

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

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

(x^{2} - x + 1) e^{3} x

y = 6 x^{2} - 3 x^{3}

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

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

y = 5 x^{8} - 6 x + 12

What is the derivative of ((3x^2+x))/((3x^2-x)) ?
The derivative of ((3x^2+x))/((3x^2-x)) is -6/((3x-1)^2)
What is the first derivative of ((3x^2+x))/((3x^2-x)) ?
The first derivative of ((3x^2+x))/((3x^2-x)) is -6/((3x-1)^2)