What is the derivative of 1/6 ln|3sec(2x-2|+C) ?

The derivative of 1/6 ln|3sec(2x-2|+C) is (sec(2x)tan(2x))/(3sec(2x)-2)

What is the first derivative of 1/6 ln|3sec(2x-2|+C) ?

The first derivative of 1/6 ln|3sec(2x-2|+C) is (sec(2x)tan(2x))/(3sec(2x)-2)

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\frac{d}{d x} (\frac{1}{6} ln ∣ 3 sec (2 x) - 2 ∣ + C)

\frac{sec ( 2 x ) tan ( 2 x )}{3 sec ( 2 x ) - 2}

Solution steps

\frac{d}{d x} (\frac{1}{6} ln ∣ 3 sec (2 x) - 2 ∣ + C)

Treat

C

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (\frac{1}{6} ln ∣ 3 sec (2 x) - 2 ∣) + \frac{d C}{d x}

\frac{d}{d x} (\frac{1}{6} ln ∣ 3 sec (2 x) - 2 ∣) = \frac{sec ( 2 x ) tan ( 2 x )}{3 sec ( 2 x ) - 2}

\frac{d C}{d x} = 0

= \frac{sec ( 2 x ) tan ( 2 x )}{3 sec ( 2 x ) - 2} + 0

Simplify

= \frac{sec ( 2 x ) tan ( 2 x )}{3 sec ( 2 x ) - 2}

\int_{0}^{\infty} \frac{1}{x + 1} d x

n = 0 \sum \infty (- 2)^{3 - 2 n} 3

\frac{\partial}{\partial x} (2 x y + 2 x)

\frac{d y}{d x} = y - x, y (0) = \frac{2}{3}

x^{^{'}} = (x^{2} - 1) t

What is the derivative of 1/6 ln|3sec(2x-2|+C) ?
The derivative of 1/6 ln|3sec(2x-2|+C) is (sec(2x)tan(2x))/(3sec(2x)-2)
What is the first derivative of 1/6 ln|3sec(2x-2|+C) ?
The first derivative of 1/6 ln|3sec(2x-2|+C) is (sec(2x)tan(2x))/(3sec(2x)-2)