find h'(x): has ln in equation

**williamb** · March 9th 2009, 06:40 PM

h(x) = −12 ln( tanx + secx ) + ln( 6 tanx ) − 8 ln( secx)
find h'(x)

alright so this is the third time doing this question, obviously i'm missing something here. i seperated each section
a) −12 ln( tanx + secx )
b) ln( 6 tanx )
c) − 8 ln( secx)

to do the chain rule on each and then combined them again into my final equation, which is wrong.
i don't know how to format text on here so try to bear with the brackets (sorry)

here's my equation:

([-12*([sec(x)]^2)+sec(x)*tan(x)]/[tan(x)+sec(x)])+([6*([sec(x)]^2)]/[6*tan(x)])-([8*sec(x)*tan(x)]/[sec(x)])

help please

thanks,
brittany

**Chris L T521** · March 9th 2009, 07:01 PM

Originally Posted by williamb

h(x) = −12 ln( tanx + secx ) + ln( 6 tanx ) − 8 ln( secx)
find h'(x)

alright so this is the third time doing this question, obviously i'm missing something here. i seperated each section
a) −12 ln( tanx + secx )
b) ln( 6 tanx )
c) − 8 ln( secx)

to do the chain rule on each and then combined them again into my final equation, which is wrong.
i don't know how to format text on here so try to bear with the brackets (sorry)

here's my equation:

([-12*[([sec(x)]^2)+sec(x)*tan(x)]]/[tan(x)+sec(x)])+([6*([sec(x)]^2)]/[6*tan(x)])-([8*sec(x)*tan(x)]/[sec(x)])

help please

thanks,
brittany

I put an additional set of brackets in your solution.

Note this can be simplified pretty nicely

$-12\frac{\sec^2x+\sec x\tan x}{\sec x+\tan x}+\frac{6\sec^2x}{6\tan x}-8\frac{\sec x\tan x}{\sec x}$ $=-12\frac{\sec x\left(\sec x+\tan x\right)}{\sec x+\tan x}+\sec^2x\cot x-8\tan x=\color{red}\boxed{-12\sec x+\sec x\csc x-8\tan x}$

Does this make sense?

w00t!!! My 2

th post!!

**Soroban** · March 9th 2009, 07:03 PM

Hello, williamb!

$h(x) \:=\:-12\ln(\tan x + \sec x) + \ln(6\tan x) - 8\ln(\sec x)$

Find $h'(x)$

$h'(x) \;=\;-12\!\cdot\!\frac{1}{\tan x + \sec x}\!\cdot\!(\sec^2\!x+\sec x\tan x) + \frac{1}{6\tan x}\!\cdot\!(6\sec^2\!x) \;-$ $8\!\cdot\!\frac{1}{\sec x}\!\cdot\!(\sec x\tan x)$

. . . . $= \;\frac{-12\sec x(\sec x + \tan x)}{\tan x + \sec x} + \frac{\sec^2\!x}{\tan x} - 8\tan x$

. . . . $= \;-12\sec x + \frac{\tan^2\!x + 1}{\tan x} - 8\tan x$

. . . . $= \;-12\sec x + \frac{\tan^2\!x}{\tan x} + \frac{1}{\tan x} - 8\tan x$

. . . . $= \;-12\sec x + \tan x + \cot x - 8\tan x$

. . . . $= \;-12\sec x - 7\tan x + \cot x$

**mollymcf2009** · March 9th 2009, 07:03 PM

Originally Posted by williamb

h(x) = −12 ln( tanx + secx ) + ln( 6 tanx ) − 8 ln( secx)
find h'(x)

alright so this is the third time doing this question, obviously i'm missing something here. i seperated each section
a) −12 ln( tanx + secx )
b) ln( 6 tanx )
c) − 8 ln( secx)

to do the chain rule on each and then combined them again into my final equation, which is wrong.
i don't know how to format text on here so try to bear with the brackets (sorry)

here's my equation:

([-12*([sec(x)]^2)+sec(x)*tan(x)]/[tan(x)+sec(x)])+([6*([sec(x)]^2)]/[6*tan(x)])-([8*sec(x)*tan(x)]/[sec(x)])

help please

thanks,
brittany

Hi Brittany,

Be careful separating it like that, trying to put it all back together and getting your signs correct might be tricky. I suggest just writing it out one step at a time but keeping it all together.

$h(x) = -12 ln(tanx + secx) + ln(6tanx) - 8ln(secx)$

*I switched the order to get that negative out of the front:

$h'(x) = \frac{1}{(6tanx)} \cdot 6sec^2x - \frac{12}{(tanx + secx)} \cdot sec^2x + secxtanx - \frac{8}{(secx)} \cdot secxtanx$

$= \frac{6 sec^2x}{6 tanx} - \frac{12 sec^2x + secxtanx}{tanx + secx} - \frac{8 secxtanx}{secx}$

From here, get a common denominator and combine like terms and then cancel. Can you take it from here?

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