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

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

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

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

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derivative of

(2 x - 1)^{2} ln (2 x - 1)

4 ln (2 x - 1) (2 x - 1) + 2 (2 x - 1)

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = (2 x - 1)^{2}, g = ln (2 x - 1)

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

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

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

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

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

= 4 ln (2 x - 1) (2 x - 1) + 2 (2 x - 1)

\int (x^{2} - 7)^{2} (2 x) d x

ln (\frac{5 x + 1}{x ^{2} - 3})

\frac{3}{x + 3}, at x = 4

y^{^{'}} = \frac{( x ^{3} + y ^{3} )}{3 x y ^{2}}

\frac{y}{x ^{2}} d x + (y - \frac{1}{x}) d y = 0

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