Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter
    Offline

    1
    ReputationRep:
    I was looking over this example on differentiating inverse functions, and I really dont get how they get the answer from the second last line:
    A function is defined as f (x) = 3e2x + 4
    Find the derivative of the inverse function.

    Solution
    Let y = f-1 (x)
    then f (y) = 3e^2y + 4 = x
    and f ' (y) = 6e^2y

    Again, using the rule for differentiating an inverse function, we have

    d/dx (f^-1(x)) = 1/f'(y)

    = 1/6e^2y

    = 1/(2x -8)

    Could anyone help (in plain English, lol)? I'm really having a bit of trouble getting to grips with this.
    Offline

    2
    ReputationRep:
    the differential of f-1(x) is 1/g'(x), where g(x) is f-1(x)

    as d/dx (f^-1(x)) = 1/f'(y) and 1/f'(y) = 1/6e^2y :

    f (f-1(x)) = x = f(y) = 3e2y + 4

    so: ln|{x - 4}/3| = 2y
    6e2y = 6eln|{x - 4}/3| = 6{x-4}/3 = 2x - 8 = f'(y)

    so: 1/f'(y) = 1/(2x -8)
    Offline

    2
    ReputationRep:
    It might help to forget about the "f"s and just use x and y.

    y = 3e^(2x) + 4

    dy/dx = 6e^(2x)

    dx/dy
    = 1 / [dy/dx]
    = 1 / (6e^(2x))
    = 1 / (2y - 8)

    Translating into function notation,

    (d/dx)[f^(-1)(x)] = 1 / (2x - 8)
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by Jonny W)
    It might help to forget about the "f"s and just use x and y.

    y = 3e^(2x) + 4

    dy/dx = 6e^(2x)

    dx/dy
    = 1 / [dy/dx]
    = 1 / (6e^(2x))
    = 1 / (2y - 8)


    Translating into function notation,

    (d/dx)[f^(-1)(x)] = 1 / (2x - 8)
    Thanks to both of you for replying so quickly. Jonny W, I'm probably being really thick here but i'm not sure about the lines in bold. how did you go from
    1 / (6e^(2x)) to 1 / (2y - 8)?

    I'm really bad at maths, sorry
    Offline

    16
    ReputationRep:
    (Original post by friendlyneutron)
    Thanks to both of you for replying so quickly. Jonny W, I'm probably being really thick here but i'm not sure about the lines in bold. how did you go from
    1 / (6e^(2x)) to 1 / (2y - 8)?

    I'm really bad at maths, sorry
    y = 3e^(2x) + 4
    => 2y = 6e^(2x) + 8
    => 6e^(2x) = 2y - 8
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by Widowmaker)
    y = 3e^(2x) + 4
    => 2y = 6e^(2x) + 8
    => 6e^(2x) = 2y - 8
    I understand that step a bit better now thanks, but how can you go from 2y - 8 to 2x - 8. I think there's something fundamental i don't really get here.
    • Thread Starter
    Offline

    1
    ReputationRep:
    Can no-one clarify it any further for me? I'd really appreciate it. I'll even resort to bribery! (i.e. rep )
    Offline

    13
    ReputationRep:
    Ohh. I like rep . He changes from y to x simply to make it look normal (it's standard to have x as the variable, not y). You've probably done it when finding inverses of functions. Say, find the inverse of y = e^x + 4. x = ln(y - 4) and so the inverse function is y = ln(x - 4) (switch x and y).
    Offline

    2
    ReputationRep:
    y = 3e^(2x) + 4

    dy/dx = 6e^(2x)

    dx/dy
    = 1 / [dy/dx]
    = 1 / (6e^(2x))

    but y = 3e^(2x) + 4, so 2y = 6e^(2x) + 8, therefore (2y - 8) = 6e^(2x)
    therefore (substituting in):
    dx/dy = 1 / (2y - 8)

    etc etc
    Offline

    16
    ReputationRep:
    (Original post by wacabac)
    y = 3e^(2x) + 4

    dy/dx = 6e^(2x)

    dx/dy
    = 1 / [dy/dx]
    = 1 / (6e^(2x))

    but y = 3e^(2x) + 4, so 2y = 6e^(2x) + 8, therefore (2y - 8) = 6e^(2x)
    therefore (substituting in):
    dx/dy = 1 / (2y - 8)

    etc etc
    How has that helped anyone?
    You've just repeated what's been posted earlier.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: April 12, 2006
Poll
Are you going to a festival?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.