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

The derivative of e^{2x-1}ln(2x-1) is 2e^{2x-1}ln(2x-1)+(2e^{2x-1})/(2x-1)

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

The first derivative of e^{2x-1}ln(2x-1) is 2e^{2x-1}ln(2x-1)+(2e^{2x-1})/(2x-1)

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

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

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

\int \frac{9}{( x ^{2} - 1 ) ^{2}} d x

d er i v a t i v e sec^{7} (x)

\frac{d y}{d x} = 2 x

\frac{d}{d x} (3 x + 3)

\frac{\partial}{\partial x} (2 e^{4 x y})

What is the derivative of e^{2x-1}ln(2x-1) ?
The derivative of e^{2x-1}ln(2x-1) is 2e^{2x-1}ln(2x-1)+(2e^{2x-1})/(2x-1)
What is the first derivative of e^{2x-1}ln(2x-1) ?
The first derivative of e^{2x-1}ln(2x-1) is 2e^{2x-1}ln(2x-1)+(2e^{2x-1})/(2x-1)