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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

(y + 4) \frac{d y}{d x} = 2 x y + 4 y

\int (cos (6 x) - csc^{2} (x)) d x

\int \frac{4 + 10 x}{7 + 4 x + 5 x ^{2}} d x

(4 + sec (θ)) sin (θ)

e^{x^{4} + 3 x}

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