What is the derivative of 5ln(sec(x+tan(x))) ?

The derivative of 5ln(sec(x+tan(x))) is -5/((sec(x)+tan(x))(-1+sin(x)))

What is the first derivative of 5ln(sec(x+tan(x))) ?

The first derivative of 5ln(sec(x+tan(x))) is -5/((sec(x)+tan(x))(-1+sin(x)))

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\frac{d}{d x} (5 ln (sec (x) + tan (x)))

- \frac{5}{( sec ( x ) + tan ( x ) ) ( - 1 + sin ( x ) )}

Solution steps

\frac{d}{d x} (5 ln (sec (x) + tan (x)))

Take the constant out:

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

= 5 \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 )}

= 5 \cdot \frac{1}{sec ( x ) + tan ( x )} (- \frac{1}{- 1 + sin ( x )})

Simplify

5 \cdot \frac{1}{sec ( x ) + tan ( x )} (- \frac{1}{- 1 + sin ( x )}) : - \frac{5}{( sec ( x ) + tan ( x ) ) ( - 1 + sin ( x ) )}

= - \frac{5}{( sec ( x ) + tan ( x ) ) ( - 1 + sin ( x ) )}

\int \frac{( arctan ( x ) ) ^{2}}{1 + x ^{2}} d x

y y^{^{'}} + x y^{2} = (e^{- x^{2}})

f (x) = \frac{1}{3} x^{3} + 5.5 x^{2} + 30 x + 22

x \to 0 + lim (\frac{x ^{x} - 1}{2 x})

t \to 0 lim (1 + t^{3})

What is the derivative of 5ln(sec(x+tan(x))) ?
The derivative of 5ln(sec(x+tan(x))) is -5/((sec(x)+tan(x))(-1+sin(x)))
What is the first derivative of 5ln(sec(x+tan(x))) ?
The first derivative of 5ln(sec(x+tan(x))) is -5/((sec(x)+tan(x))(-1+sin(x)))