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

The derivative of ln((x^4/((3x-4)^2))) is (2(3x-8))/(x(3x-4))

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

The first derivative of ln((x^4/((3x-4)^2))) is (2(3x-8))/(x(3x-4))

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

\frac{2 ( 3 x - 8 )}{x ( 3 x - 4 )}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{x ^{4}}{( 3 x - 4 ) ^{2}}) = \frac{2 x ^{3} ( 3 x - 8 )}{( 3 x - 4 ) ^{3}}

= \frac{1}{\frac{x ^{4}}{( 3 x - 4 ) ^{2}}} \cdot \frac{2 x ^{3} ( 3 x - 8 )}{( 3 x - 4 ) ^{3}}

Simplify

\frac{1}{\frac{x ^{4}}{( 3 x - 4 ) ^{2}}} \cdot \frac{2 x ^{3} ( 3 x - 8 )}{( 3 x - 4 ) ^{3}} : \frac{2 ( 3 x - 8 )}{x ( 3 x - 4 )}

= \frac{2 ( 3 x - 8 )}{x ( 3 x - 4 )}

d er i v a t i v e 2 π x^{2}

y^{^{'}} + \frac{1}{200 + t} y = 0

x \to 1 + lim (x^{2} - 1)

d er i v a t i v e f (x) = ln (ln (4 x))

\frac{d x}{d y} = \frac{y - x}{y + x}

What is the derivative of ln((x^4/((3x-4)^2))) ?
The derivative of ln((x^4/((3x-4)^2))) is (2(3x-8))/(x(3x-4))
What is the first derivative of ln((x^4/((3x-4)^2))) ?
The first derivative of ln((x^4/((3x-4)^2))) is (2(3x-8))/(x(3x-4))