What is the derivative of 7ln(sec(x+tan(x))) ?

The derivative of 7ln(sec(x+tan(x))) is -7/((sec(x)+tan(x))(-1+sin(x)))

What is the first derivative of 7ln(sec(x+tan(x))) ?

The first derivative of 7ln(sec(x+tan(x))) is -7/((sec(x)+tan(x))(-1+sin(x)))

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\frac{d}{d x} (7 ln (sec (x) + tan (x)))

- \frac{7}{( sec ( x ) + tan ( x ) ) ( - 1 + sin ( x ) )}

Solution steps

\frac{d}{d x} (7 ln (sec (x) + tan (x)))

Take the constant out:

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

= 7 \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 )}

= 7 \cdot \frac{1}{sec ( x ) + tan ( x )} (- \frac{1}{- 1 + sin ( x )})

Simplify

7 \cdot \frac{1}{sec ( x ) + tan ( x )} (- \frac{1}{- 1 + sin ( x )}) : - \frac{7}{( sec ( x ) + tan ( x ) ) ( - 1 + sin ( x ) )}

= - \frac{7}{( sec ( x ) + tan ( x ) ) ( - 1 + sin ( x ) )}

x y^{^{'}} + y = 26 x

y = 2 x^{3} - 6 x^{2} + 5 x + 1, (2, 3)

h^{^{'}} = - h

\frac{d}{d x} (\frac{4}{3})

\frac{d}{d x} (sin^{3} (x) cos (x))

What is the derivative of 7ln(sec(x+tan(x))) ?
The derivative of 7ln(sec(x+tan(x))) is -7/((sec(x)+tan(x))(-1+sin(x)))
What is the first derivative of 7ln(sec(x+tan(x))) ?
The first derivative of 7ln(sec(x+tan(x))) is -7/((sec(x)+tan(x))(-1+sin(x)))