What is the derivative of (5x/(ln((4x-1)^2))) ?

The derivative of (5x/(ln((4x-1)^2))) is (5(ln(4x-1)(4x-1)-4x))/(2ln^2(4x-1)(4x-1))

What is the first derivative of (5x/(ln((4x-1)^2))) ?

The first derivative of (5x/(ln((4x-1)^2))) is (5(ln(4x-1)(4x-1)-4x))/(2ln^2(4x-1)(4x-1))

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\frac{d}{d x} \frac{5 x}{ln ( ( 4 x - 1 ) ^{2} )}

\frac{5 ( ln ( 4 x - 1 ) ( 4 x - 1 ) - 4 x )}{2 ln ^{2} ( 4 x - 1 ) ( 4 x - 1 )}

Solution steps

\frac{d}{d x} \frac{5 x}{ln ( ( 4 x - 1 ) ^{2} )}

Take the constant out:

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

= 5 \frac{d}{d x} \frac{x}{ln ( ( 4 x - 1 ) ^{2} )}

Apply the Quotient Rule:

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

= 5 \cdot \frac{\frac{d x}{d x} ln ( ( 4 x - 1 ) ^{2} ) - \frac{d}{d x} ( ln ( ( 4 x - 1 ) ^{2} ) ) x}{( ln ( ( 4 x - 1 ) ^{2} ) ) ^{2}}

\frac{d x}{d x} = 1

\frac{d}{d x} (ln ((4 x - 1)^{2})) = \frac{8}{4 x - 1}

= 5 \cdot \frac{1 \cdot ln ( ( 4 x - 1 ) ^{2} ) - \frac{8}{4 x - 1} x}{( ln ( ( 4 x - 1 ) ^{2} ) ) ^{2}}

Simplify

5 \cdot \frac{1 \cdot ln ( ( 4 x - 1 ) ^{2} ) - \frac{8}{4 x - 1} x}{ln ^{2} ( ( 4 x - 1 ) ^{2} )} : \frac{5 ( ln ( 4 x - 1 ) ( 4 x - 1 ) - 4 x )}{2 ln ^{2} ( 4 x - 1 ) ( 4 x - 1 )}

= \frac{5 ( ln ( 4 x - 1 ) ( 4 x - 1 ) - 4 x )}{2 ln ^{2} ( 4 x - 1 ) ( 4 x - 1 )}

x \to \infty lim (g (x))

5 x^{2} - 3 x - 5

f (x) = e^{x^{7}}, (- 1, \frac{1}{e})

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

\int_{0}^{\frac{π}{2}} 7 cos^{5} (x) d x

What is the derivative of (5x/(ln((4x-1)^2))) ?
The derivative of (5x/(ln((4x-1)^2))) is (5(ln(4x-1)(4x-1)-4x))/(2ln^2(4x-1)(4x-1))
What is the first derivative of (5x/(ln((4x-1)^2))) ?
The first derivative of (5x/(ln((4x-1)^2))) is (5(ln(4x-1)(4x-1)-4x))/(2ln^2(4x-1)(4x-1))