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

    15
    ReputationRep:
    Find the gradient of the curve with equation y=f(x) at the point A where:

    f(x)=x^3-3x+2 and A is at (-1,4)

    This is what I've done so far but I think it's wrong.

    Name:  Eqn2.gif
Views: 38
Size:  906 Bytes
    • Study Helper
    Offline

    7
    ReputationRep:
    (Original post by zed963)
    Find the gradient of the curve with equation y=f(x) at the point A where:

    f(x)=x^3-3x+2 and A is at (-1,4)

    This is what I've done so far but I think it's wrong.

    Name:  Eqn2.gif
Views: 38
Size:  906 Bytes
    You have gone wrong twice, First with your differentiation. and why have you differentiated twice !!
     \frac{d}{dx}(3x)=3 so  \frac{dy}{dx}=3x^2-3
    x=-1 \implies  3x^2=3 so final answer should be 0
    Offline

    2
    ReputationRep:
    (Original post by zed963)
    Find the gradient of the curve with equation y=f(x) at the point A where:

    f(x)=x^3-3x+2 and A is at (-1,4)

    This is what I've done so far but I think it's wrong.

    Name:  Eqn2.gif
Views: 38
Size:  906 Bytes
    Like the above poster ^ said, EVERY term gets differentiated. So -3x and 2 also get differentiated. Finding out dy/dx means finding an equation that allows you to work out the curve's gradient at a point, which is what you need to do just once. But differentiating a further time means you finding an equation that allows you to determine at a point the rate of change of the gradient. This is not what you want, so you shouldn't have differentiated the extra time.
    Offline

    2
    ReputationRep:
    Also I forgot to mention that if you have a curve's equation and you differentiate it, you get dy/dx. Differentiating one more time gives you \frac{d^2 y}{dx^2}. Differentiating again gives \frac{d^3 y}{dx^3}. Differentiating another time gives you \frac{d^4 y}{dx^4} etc...
    • Study Helper
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by krisshP)
    Also I forgot to mention that if you have a curve's equation and you differentiate it, you get dy/dx. Differentiating one more time gives you \frac{d^2 y}{dx^2}. Differentiating again gives \frac{d^3 y}{dx^3}. Differentiating another time gives you \frac{d^4 y}{dx^4} etc...

    (Original post by brianeverit)
    You have gone wrong twice, First with your differentiation. and why have you differentiated twice !!
     \frac{d}{dx}(3x)=3 so  \frac{dy}{dx}=3x^2-3
    x=-1 \implies  3x^2=3 so final answer should be 0
    Oh okay,

    What I've done now.

    \ f(x)=x^3-3x+2

    \frac{dy}{dx}=3x^2-3+2=3x^2-1

    \ 3(-1)^2-1=2
    Offline

    2
    ReputationRep:
    (Original post by zed963)
    Oh okay,

    What I've done now.

    \ f(x)=x^3-3x+2

    \frac{dy}{dx}=3x^2-3+2=3x^2-1

    \ 3(-1)^2-1=2
    NO!

    The 2 is differentiated to 0. Look

    2
    =2x^0
    =0 X 2x^-1

    0 x anything=0

    so differentiating 2 gives 0.
    • Study Helper
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by krisshP)
    NO!

    The 2 is differentiated to 0. Look

    2
    =2x^0
    =0 X 2x^-1

    0 x anything=0

    so differentiating 2 gives 0.
    Oh...

    Yes that's true.

    \ 3x^2-3

    \ 3(-1)^2-3=0
    • Study Helper
    • Thread Starter
    Offline

    15
    ReputationRep:
    (Original post by krisshP)
    NO!

    The 2 is differentiated to 0. Look

    2
    =2x^0
    =0 X 2x^-1

    0 x anything=0

    so differentiating 2 gives 0.
    The two is on its own, why have you included an x?
    Offline

    2
    ReputationRep:
    (Original post by zed963)
    The two is on its own, why have you included an x?
    Because it allows us to see clearly and easily how we can differentiate it.
 
 
 
  • 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.