What is the derivative of 6ln(sec(x+tan(x))) ?

The derivative of 6ln(sec(x+tan(x))) is -6/((sec(x)+tan(x))(-1+sin(x)))

What is the first derivative of 6ln(sec(x+tan(x))) ?

The first derivative of 6ln(sec(x+tan(x))) is -6/((sec(x)+tan(x))(-1+sin(x)))

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\frac{d}{d x} (6 ln (sec (x) + tan (x)))

- \frac{6}{( sec ( x ) + tan ( x ) ) ( - 1 + sin ( x ) )}

Solution steps

\frac{d}{d x} (6 ln (sec (x) + tan (x)))

Take the constant out:

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

= 6 \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 )}

= 6 \cdot \frac{1}{sec ( x ) + tan ( x )} (- \frac{1}{- 1 + sin ( x )})

Simplify

6 \cdot \frac{1}{sec ( x ) + tan ( x )} (- \frac{1}{- 1 + sin ( x )}) : - \frac{6}{( sec ( x ) + tan ( x ) ) ( - 1 + sin ( x ) )}

= - \frac{6}{( sec ( x ) + tan ( x ) ) ( - 1 + sin ( x ) )}

cos^{3} (x)

y^{^{'}} - \frac{6}{x} y = \frac{y ^{4}}{x ^{15}}

- 6 sin (t)

ln (e^{x} + x)

4 x y^{^{'}} + y = 20 x

What is the derivative of 6ln(sec(x+tan(x))) ?
The derivative of 6ln(sec(x+tan(x))) is -6/((sec(x)+tan(x))(-1+sin(x)))
What is the first derivative of 6ln(sec(x+tan(x))) ?
The first derivative of 6ln(sec(x+tan(x))) is -6/((sec(x)+tan(x))(-1+sin(x)))