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

The derivative of ln(((2x+1^3)/((3x-1)^4))) is (6(-x-3))/((2x+1)(3x-1))

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

The first derivative of ln(((2x+1^3)/((3x-1)^4))) is (6(-x-3))/((2x+1)(3x-1))

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

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

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{( 2 x + 1 ) ^{3}}{( 3 x - 1 ) ^{4}}) = \frac{6 ( 2 x + 1 ) ^{2} ( - x - 3 )}{( 3 x - 1 ) ^{5}}

= \frac{1}{\frac{( 2 x + 1 ) ^{3}}{( 3 x - 1 ) ^{4}}} \cdot \frac{6 ( 2 x + 1 ) ^{2} ( - x - 3 )}{( 3 x - 1 ) ^{5}}

Simplify

\frac{1}{\frac{( 2 x + 1 ) ^{3}}{( 3 x - 1 ) ^{4}}} \cdot \frac{6 ( 2 x + 1 ) ^{2} ( - x - 3 )}{( 3 x - 1 ) ^{5}} : \frac{6 ( - x - 3 )}{( 2 x + 1 ) ( 3 x - 1 )}

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

z (z + 1) e^{z}

f (x) = 16 - x^{2}, (1, 4)

\frac{\partial}{\partial x} ((2 x y - 1) e^{- x y})

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

(9 + y^{2}) d x + 4 x y d y = 0

What is the derivative of ln(((2x+1^3)/((3x-1)^4))) ?
The derivative of ln(((2x+1^3)/((3x-1)^4))) is (6(-x-3))/((2x+1)(3x-1))
What is the first derivative of ln(((2x+1^3)/((3x-1)^4))) ?
The first derivative of ln(((2x+1^3)/((3x-1)^4))) is (6(-x-3))/((2x+1)(3x-1))