What is the derivative of 3ln(sec(x+tan(x))) ?

The derivative of 3ln(sec(x+tan(x))) is -3/((sec(x)+tan(x))(-1+sin(x)))

What is the first derivative of 3ln(sec(x+tan(x))) ?

The first derivative of 3ln(sec(x+tan(x))) is -3/((sec(x)+tan(x))(-1+sin(x)))

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\frac{d}{d x} (3 ln (sec (x) + tan (x)))

- \frac{3}{( sec ( x ) + tan ( x ) ) ( - 1 + sin ( x ) )}

Solution steps

\frac{d}{d x} (3 ln (sec (x) + tan (x)))

Take the constant out:

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

= 3 \frac{d}{d x} (ln (sec (x) + tan (x)))

Apply the chain rule:

\frac{1}{sec ( x ) + tan ( x )} \frac{d}{d x} (sec (x) + tan (x))

= \frac{1}{sec ( x ) + tan ( x )} \frac{d}{d x} (sec (x) + tan (x))

\frac{d}{d x} (sec (x) + tan (x)) = - \frac{1}{- 1 + sin ( x )}

= 3 \cdot \frac{1}{sec ( x ) + tan ( x )} (- \frac{1}{- 1 + sin ( x )})

Simplify

3 \cdot \frac{1}{sec ( x ) + tan ( x )} (- \frac{1}{- 1 + sin ( x )}) : - \frac{3}{( sec ( x ) + tan ( x ) ) ( - 1 + sin ( x ) )}

= - \frac{3}{( sec ( x ) + tan ( x ) ) ( - 1 + sin ( x ) )}

y^{^{'}} - e^{x + 1} = 0

\int sec (\frac{x}{4}) d x

y^{^{''}} + 9 y = 2 sin (x) + cos (x)

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

25 y^{^{''}} + 90 y^{^{'}} + 81 y = 0

What is the derivative of 3ln(sec(x+tan(x))) ?
The derivative of 3ln(sec(x+tan(x))) is -3/((sec(x)+tan(x))(-1+sin(x)))
What is the first derivative of 3ln(sec(x+tan(x))) ?
The first derivative of 3ln(sec(x+tan(x))) is -3/((sec(x)+tan(x))(-1+sin(x)))