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

Maclaurin's theorem watch

Announcements
    • Thread Starter
    Offline

    1
    ReputationRep:
    I was doing a question - Given that y = ln cos(2X), find d^4y/dX^4

    I got dy/dx as -2sin2X/cos2X and d^2y/dX^2 as -4/cos^2(2X) (the markscheme put -4sec^2(2X) which I'm pretty sure is the same)

    Then for d^3y/dX^3, I used the quotient rule to get d^3y/dX^3 =

    cos(2X)(0) - -4(d/dX(cos^2(2X))) / (cos^2(2X))^2

    for d/dX(cos^2(2X)), I put it equal to d/dX ((cos2X)^2) = -4sin(2X) x cos(2X)

    Then I got d^3y/dX^3 = -16sin(2X) x (cos(2X)) / (cos(2X))^2, so I cancelled out the cos(2X) to get d^3y / dX^3 = -16sin(2X) / cos(2X) = -16tan(2X)

    But the markscheme says it is -16tan(2X) x sec^2(2X)

    Where have I gone wrong please???
    • Political Ambassador
    Offline

    3
    ReputationRep:
    Political Ambassador
    (Original post by MathsALevel123)
    I was doing a question - Given that y = ln cos(2X), find d^4y/dX^4

    I got dy/dx as -2sin2X/cos2X and d^2y/dX^2 as -4/cos^2(2X) (the markscheme put -4sec^2(2X) which I'm pretty sure is the same)

    Then for d^3y/dX^3, I used the quotient rule to get d^3y/dX^3 =

    cos(2X)(0) - -4(d/dX(cos^2(2X))) / (cos^2(2X))^2

    for d/dX(cos^2(2X)), I put it equal to d/dX ((cos2X)^2) = -4sin(2X) x cos(2X)

    Then I got d^3y/dX^3 = -16sin(2X) x (cos(2X)) / (cos(2X))^2, so I cancelled out the cos(2X) to get d^3y / dX^3 = -16sin(2X) / cos(2X) = -16tan(2X)

    But the markscheme says it is -16tan(2X) x sec^2(2X)

    Where have I gone wrong please???
    You just had a bit of a "mental block"

    (\cos^2 (2x))^2 = \cos^4 (2x)

    but you put \cos^2 2x

    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by Indeterminate)
    You just had a bit of a "mental block"

    (\cos^2 (2x))^2 = \cos^4 (2x)

    but you put \cos^2 2x

    Thanks!!!
    • Political Ambassador
    Offline

    3
    ReputationRep:
    Political Ambassador
    (Original post by MathsALevel123)
    Thanks!!!
    My pleasure

    Besides we've all made mistakes like that :lol:
 
 
 
Poll
Do you like carrot cake?

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.