Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    What is the 2nd derivative of :
    tan^2(2x^2 +3)

    Thanks
    Offline

    0
    ReputationRep:
    (Original post by kevins)
    What is the 2nd derivative of :
    tan^2(2x^2 +3)

    Thanks
    The working is quite tedious, so I'll simply state the derivatives;

    y = (tan(2x² +3))²

    dy/dx = 8xsec(2x²+3)²tan(2x² +3)

    d²y/dx² = -4sec(2x²+3)^4.[8cos(4x² +6)x²-16x²-sin(4x² +6)]
    Offline

    0
    ReputationRep:
    8x.sec(2x²+3)².tan(2x²+3)
    Offline

    0
    ReputationRep:
    (Original post by manps)
    8x.sec(2x²+3)².tan(2x²+3)
    He wants d²y/dx²
    Offline

    18
    ReputationRep:
    (Original post by manps)
    8x.sec(2x²+3)².tan(2x²+3)
    2nd derivative dude.
    • Thread Starter
    Offline

    0
    ReputationRep:
    [QUOTE]
    Thanks but can u put it in terms on sec and tan. really appreciate it
    Offline

    0
    ReputationRep:
    [QUOTE=kevins]
    Thanks but can u put it in terms on sec and tan. really appreciate it
    Now you are really asking a difficult question.
    Offline

    18
    ReputationRep:
    (Original post by Dekota)
    Now you are really asking a difficult question.


    Some people are so lazy. :rolleyes:
    • Thread Starter
    Offline

    0
    ReputationRep:
    its jsut tha wheni do it i get my answer in terms of tan and sec and yours has sin and cos in it
    Offline

    0
    ReputationRep:
    (Original post by kevins)
    its jsut tha wheni do it i get my answer in terms of tan and sec and yours has sin and cos in it
    You do one thing, to check if yours is the same as mine...

    Let x = 1, and substitute this into your answer for d²y/dx²

    and then let x = 1, and substitute this into my answer for d²y/dx²...
    d²y/dx² = -4sec(2x²+3)^4.[8cos(4x² +6)x²-16x²-sin(4x² +6)]

    If both give the same answer, then obviously they are equal and the same.

    But I will say this; your derivative (answer) seems wrong. :cool:
    • Thread Starter
    Offline

    0
    ReputationRep:
    yeh yours seems to be right def, thanks alot but would i be pushing it if i was askin for the working out. i can give you my email adress if you want.
    Offline

    0
    ReputationRep:
    (Original post by kevins)
    yeh yours seems to be right def, thanks alot but would i be pushing it if i was askin for the working out. i can give you my email adress if you want.
    How about, tomorrow .. I'll post it up?
    • Thread Starter
    Offline

    0
    ReputationRep:
    yeh thanks mate
    Offline

    0
    ReputationRep:
    (Original post by kevins)
    What is the 2nd derivative of :
    tan^2(2x^2 +3)
    y = [tan(2x² +3)]²
    dy/dx = 2[tan(2x² +3)].d[tan(2x² +3)]/dx
    .........= 2[tan(2x² +3)].4xsec(2x² +3)²
    .........= 8x[tan(2x² +3)]sec(2x² +3)²

    1/8.d²y/dx²....= x[tan(2x² +3)]sec(2x² +3)²
    ...................= x[sin(2x² +3)/cos(2x² +3)]sec(2x² +3)²
    ...................= xsin(2x² +3)/[cos(2x² +3)cos(2x² +3)²]
    ...................= d[xsin(2x² +3)]/dx.[cos(2x² +3)cos(2x² +3)²] - xsin(2x² +3)].d[cos(2x² +3)cos(2x² +3)²]/dx
    ...................= [x.d[sin(2x² +3)]/dx + d[x]/dx.sin(2x² +3)].[cos(2x² +3)cos(2x² +3)²] - xsin(2x² +3)].cos(2x² +3).d[cos(2x² +3)²]/dx + [d[cos(2x² +3)]/dx.cos(2x² +3)²]
    ...................= [4x²cos(2x² +3) + sin(2x² +3)].[cos(2x² +3)cos(2x² +3)²] - xsin(2x² +3)].[cos(2x² +3).-(16x³+24x)sin(4x^4 +12x²+6) - 4xsin(2x² +3)cos(2x² +3)²]
    ...................= [4x²cos(2x² +3) + sin(2x² +3)cos(2x² +3)cos(2x² +3)²] + [xsin(2x² +3)cos(2x² +3)sin(4x^4 +12x²+6) - 4xsin(2x² +3)cos(2x² +3)²]
    ...................= [[4x²cos(2x² +3) + sin(2x² +3)cos(2x² +3)cos(2x² +3)²] + [xsin(2x² +3)cos(2x² +3)sin(4x^4 +12x²+6) - 4xsin(2x² +3)cos(2x² +3)²]] / [cos(2x² +3)cos(2x² +3)²]²
    ...................= [[4x²cos(2x² +3) + sin(2x² +3)cos(2x² +3)cos(2x² +3)²] + [xsin(2x² +3)cos(2x² +3)sin(4x^4 +12x²+6) - 4xsin(2x² +3)cos(2x² +3)²]] / [cos(2x² +3)cos(2x² +3)²]²
    ...................= 8[[[4x²cos(2x² +3) + sin(2x² +3)cos(2x² +3)cos(2x² +3)²] + [xsin(2x² +3)cos(2x² +3)sin(4x^4 +12x²+6) - 4xsin(2x² +3)cos(2x² +3)²]] / [cos(2x² +3)cos(2x² +3)²]²]

    This result is from the quotient rule. I may have made mistakes and haven't checked it yet. But after clearing this expression.. you should end up with my earlier stated expression for d²y/dx², or something similar.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: November 12, 2005

University open days

  • University of Bradford
    All faculties Undergraduate
    Wed, 21 Nov '18
  • Buckinghamshire New University
    All Faculties Postgraduate
    Wed, 21 Nov '18
  • Heriot-Watt University
    All Schools Postgraduate
    Wed, 21 Nov '18
Poll
Black Friday: Yay or Nay?
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

Equations

Best calculators for A level Maths

Tips on which model to get

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.