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

The derivative of y=ln(sec(5x)+tan(5x)) is 5sec(5x)

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

The first derivative of y=ln(sec(5x)+tan(5x)) is 5sec(5x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

derivative of

y = ln (sec (5 x) + tan (5 x))

5 sec (5 x)

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (sec (5 x) + tan (5 x)) = sec (5 x) tan (5 x) \cdot 5 + sec^{2} (5 x) \cdot 5

= \frac{1}{sec ( 5 x ) + tan ( 5 x )} (sec (5 x) tan (5 x) \cdot 5 + sec^{2} (5 x) \cdot 5)

Simplify

\frac{1}{sec ( 5 x ) + tan ( 5 x )} (sec (5 x) tan (5 x) \cdot 5 + sec^{2} (5 x) \cdot 5) : 5 sec (5 x)

= 5 sec (5 x)

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