Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    3
    ReputationRep:
    Hi, I'm not sure how this question is meant to be worked out. Could somebody show me how you'd approach this?

    Find the derived function, given that f(x) equals:

    \frac{1}{x^2}
    Offline

    1
    ReputationRep:
    Reciprocal rule?
    Offline

    0
    ReputationRep:
    (Original post by Mr Inquisitive)
    Hi, I'm not sure how this question is meant to be worked out. Could somebody show me how you'd approach this?

    Find the derived function, given that f(x) equals:

    \frac{1}{x^2}
    that is the same as x^(-2)
    Can you do it now?
    Offline

    2
    ReputationRep:
    (Original post by Mr Inquisitive)
    Hi, I'm not sure how this question is meant to be worked out. Could somebody show me how you'd approach this?

    Find the derived function, given that f(x) equals:

    \frac{1}{x^2}
    1 over anything equals the denominator to the negative power. For example: \frac1{x^4} is the same as x^{-4}. So \frac{dy}{dx} or f'(x) for that example is -4x^{-5}. Hope this helps.
    • Thread Starter
    Offline

    3
    ReputationRep:
    Ah, I get it now. So it would be -2x^{-3}?
    Offline

    1
    ReputationRep:
    (Original post by Mr Inquisitive)
    Ah, I get it now. So it would be -2x^{-3}?
    Yeah thats fine
    Offline

    2
    ReputationRep:
    Yes, -2x^-3. Which can also be written as -2/x^3
    Offline

    0
    ReputationRep:
    (Original post by Mr Inquisitive)
    Ah, I get it now. So it would be -2x^{-3}?
    yeh
    You could write it back as a fraction to tidy up though
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by Rational Paradox)
    Yeah thats fine
    Are you sure? It's not being integrated.
    Offline

    2
    ReputationRep:
    Add one to the power, don't take away one from the power
    No, you should take one away from the power when differentiating.
    Offline

    2
    ReputationRep:
    (Original post by Rational Paradox)
    Yeah thats fine
    the new power is n-1 for differentiation. n+1 for integration, where n is the original power.
    Offline

    2
    ReputationRep:
    f(x) can be written and x^(-2)

    So as the rule goes, bring the power to the front, take one off the power

    -2x^(-3)

    which is the same as
    -2
    x^3
    • Thread Starter
    Offline

    3
    ReputationRep:
    Thank you all.
    Offline

    2
    ReputationRep:
    Whenever I'm faced with something like that to differentiate, I turn it back into a fraction once I've differentiated because it looks tidier. -2x^{-3} is the same as -\frac{2}{x^3}
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Did TEF Bronze Award affect your UCAS choices?
    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
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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

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