# Derivative of trig functions under the natural log

• Feb 17th 2011, 05:42 PM
bennettb6
Derivative of trig functions under the natural log

f(x)=ln(sec(6x)+tan(6x))

Explanation would be very appreciated.
• Feb 17th 2011, 05:49 PM
e^(i*pi)
Standard Chain Rule: $\dfrac{d}{dx} \ln[f(x)] = \dfrac{f'(x)}{f(x)}$

In this case you have $f'(x) = \dfrac{\frac{d}{dx} \sec(6x) + \frac{d}{dx} \tan(6x)}{\sec(6x) + \tan(6x)}$

You'll need to use the chain rule again on the numerator
• Feb 17th 2011, 05:57 PM
Soroban
Hello, bennettb6!

Quote:

$\text{Find the first and second derivative of the following function:}$

. . $f(x)\:=\:\ln(\sec6x+\tan6x)$

Chain Rule: . $\displaystyle f'(x) \;=\;\frac{1}{\sec6x + \tan6x}\cdot(\sec6x\tan6x\cdot 6 + \sec^2\!6x\cdot 6)$

. . . . . . . . . $\displaystyle f'(x) \;=\;\frac{6\sec6x\,(\tan6x + \sec6x)}{\sec6x + \tan6x} \;=\;6\sec6x$

$\text{Then: }\;f''(x) \;=\;6\sec6x\tan6x\cdot 6 \;=\;36\sec6x\tan6x$

• Feb 17th 2011, 06:12 PM
bennettb6
Perfect, thanks for the help!