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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\int (x + 3)^{3} d x

x \to - 4 lim (x^{3} + 4 x^{2} - 8 x)

y ln (y) d x + (x - ln (y)) d y = 0

\frac{\partial}{\partial x} (x^{3} y + e^{x y^{2}})

\frac{d y}{d x} = 1 - 2 y

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