Results 1 to 4 of 4

Math Help - find h'(x): has ln in equation

  1. #1
    Junior Member
    Joined
    Oct 2008
    Posts
    34

    find h'(x): has ln in equation

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by williamb View Post
    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 2th post!!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,682
    Thanks
    614
    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

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member mollymcf2009's Avatar
    Joined
    Jan 2009
    From
    Charleston, SC
    Posts
    490
    Awards
    1
    Quote Originally Posted by williamb View Post
    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?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: May 18th 2011, 10:36 AM
  2. Replies: 8
    Last Post: March 22nd 2011, 04:57 PM
  3. Find the equation
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: March 11th 2010, 07:22 AM
  4. Replies: 5
    Last Post: October 13th 2008, 10:16 AM
  5. find an equation to r(t)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 21st 2007, 10:16 PM

Search Tags


/mathhelpforum @mathhelpforum