What is the d/(dt)((x)(ln(2x-1))) ?

The d/(dt)((x)(ln(2x-1))) is ln(2t-1)+(2t)/(2t-1)

What is the first d/(dt)((x)(ln(2x-1))) ?

The first d/(dt)((x)(ln(2x-1))) is ln(2t-1)+(2t)/(2t-1)

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\frac{d}{d t} ((t) (ln (2 t - 1)))

ln (2 t - 1) + \frac{2 t}{2 t - 1}

Solution steps

\frac{d}{d t} ((t) (ln (2 t - 1)))

Apply the Product Rule:

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

f = t, g = ln (2 t - 1)

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

\frac{d t}{d t} = 1

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

= 1 \cdot ln (2 t - 1) + \frac{2}{2 t - 1} t

Simplify

1 \cdot ln (2 t - 1) + \frac{2}{2 t - 1} t : ln (2 t - 1) + \frac{2 t}{2 t - 1}

= ln (2 t - 1) + \frac{2 t}{2 t - 1}

cos^{2} (x) - 3 cos (x) + 2

64 x^{6} - 125 y^{3} z^{9}

e (- x^{2})

3 a x - 3 a y - 5 b x + 5 b y

m^{11} - 7 m^{5}

What is the d/(dt)((x)(ln(2x-1))) ?
The d/(dt)((x)(ln(2x-1))) is ln(2t-1)+(2t)/(2t-1)
What is the first d/(dt)((x)(ln(2x-1))) ?
The first d/(dt)((x)(ln(2x-1))) is ln(2t-1)+(2t)/(2t-1)