Could someone please help me figure out the first and second derivative of the following function:
f(x)=ln(sec(6x)+tan(6x))
Explanation would be very appreciated.
Standard Chain Rule: $\displaystyle \dfrac{d}{dx} \ln[f(x)] = \dfrac{f'(x)}{f(x)}$
In this case you have $\displaystyle f'(x) = \dfrac{\frac{d}{dx} \sec(6x) + \frac{d}{dx} \tan(6x)}{\sec(6x) + \tan(6x)}$
You'll need to use the chain rule again on the numerator
Hello, bennettb6!
$\displaystyle \text{Find the first and second derivative of the following function:}$
. . $\displaystyle f(x)\:=\:\ln(\sec6x+\tan6x)$
Chain Rule: .$\displaystyle \displaystyle f'(x) \;=\;\frac{1}{\sec6x + \tan6x}\cdot(\sec6x\tan6x\cdot 6 + \sec^2\!6x\cdot 6) $
. . . . . . . . .$\displaystyle \displaystyle f'(x) \;=\;\frac{6\sec6x\,(\tan6x + \sec6x)}{\sec6x + \tan6x} \;=\;6\sec6x$
$\displaystyle \text{Then: }\;f''(x) \;=\;6\sec6x\tan6x\cdot 6 \;=\;36\sec6x\tan6x$