Results are out! Find what you need...fast. Get quick advice or join the chat
x

Unlock these great extras with your FREE membership

  • One-on-one advice about results day and Clearing
  • Free access to our personal statement wizard
  • Customise TSR to suit how you want to use it

need help on dy/dx

Announcements Posted on
Rate your uni — help us build a league table based on real student views 19-08-2015
  1. Offline

    ReputationRep:
    (Original post by Mr Inquisitive)
    When y=a^n, \frac{dy}{dx}=na^{n-1} - just apply that to the y=x^3 and substitute the values of x for the gradients you require.
    is it x=3 graient =27
    x=5 gradient =75

    x=-3 gradient =27??
  2. Offline

    ReputationRep:
    (Original post by jetskiwavedunkno)
    is it x=3 graient =27
    x=5 gradient =75

    x=-3 gradient =27??
    Those are all right.
  3. Offline

    ReputationRep:
    (Original post by raheem94)
    Only the gradient at x=3 is correct, others are wrong, try again.

    How did you differentiated x^3? maybe you are making a mistake there.
    you are wrong
  4. Offline

    ReputationRep:
    (Original post by Gemini92)
    Those are all right.
    awww thank you! xx
  5. Offline

    ReputationRep:
    (Original post by jetskiwavedunkno)
    is it x=3 graient =27
    x=5 gradient =75

    x=-3 gradient =27??
    Yes, they're correct.
  6. Offline

    ReputationRep:
    (Original post by raheem94)
    Only the gradient at x=3 is correct, others are wrong, try again.
    I'm afraid the gradient at 3 is also incorrect?

    3x^2 when x = 3 gives 3 multiplied by 3^2, which is 3 x 9 = 27..

    Ooops, too late! You've got it now anyway.
  7. Offline

    ReputationRep:
    (Original post by Mr Inquisitive)
    Yes, they're correct.
    for each curve find the gradient at the given point.

    y=4x^2+2 at x=1 gradient=

    y= x^3-2 at x=1 gradient=

    y= 4x^3 at x=1 gradient=

    how do i do this
  8. Offline

    ReputationRep:
    (Original post by jetskiwavedunkno)
    for each curve find the gradient at the given point.

    y=4x^2+2 at x=1 gradient=

    y= x^3-2 at x=1 gradient=

    y= 4x^3 at x=1 gradient=

    how do i do this
    Exactly the same way as the others, differentiate the function and sub in the x value. (Remember that constants differentiate to 0)
  9. Offline

    ReputationRep:
    (Original post by jetskiwavedunkno)
    for each curve find the gradient at the given point.

    y=4x^2+2 at x=1 gradient=

    y= x^3-2 at x=1 gradient=

    y= 4x^3 at x=1 gradient=

    how do i do this
    the same way as you did before differentiate each equation, except this time you only need to sub in one x value for each equation, which is 1
  10. Offline

    ReputationRep:
    (Original post by Gemini92)
    Exactly the same way as the others, differentiate the function and sub in the x value. (Remember that constants differentiate to 0)


    y=4x^2+2 at x=1 gradient=10??
  11. Offline

    ReputationRep:
    (Original post by mr tim)
    the same way as you did before differentiate each equation, except this time you only need to sub in one x value for each equation, which is 1
    for each curve find the gradient at the given point.

    y=4x^2+2 at x=1 gradient=10??
  12. Offline

    ReputationRep:
    (Original post by jetskiwavedunkno)
    for each curve find the gradient at the given point.

    y=4x^2+2 at x=1 gradient=10??
    :no: remember that the constant [ie: number without the x next it] differentiated gives 0.
  13. Offline

    ReputationRep:
    (Original post by mr tim)
    :no: remember that the constant [ie: number without the x next it] differentiated gives 0.
    for each curve find the gradient at the given point.

    y=4x^2+2 at x=1 gradient= 8

    y= x^3-2 at x=1 gradient= 3

    y= 4x^3 at x=1 gradient=12

    are they right
  14. Offline

    ReputationRep:
    (Original post by jetskiwavedunkno)
    for each curve find the gradient at the given point.

    y=4x^2+2 at x=1 gradient=

    y= x^3-2 at x=1 gradient=

    y= 4x^3 at x=1 gradient=

    how do i do this
    Take \frac{dy}{dx} of the y value, set the value you obtain equal to zero, then substitute in x=1.

    Spoiler:
    Show
    First one would be \frac{dy}{dx}=8x, so follow on from there.
  15. Offline

    ReputationRep:
    (Original post by Mr Inquisitive)
    Take \frac{dy}{dx} of the y value, set the value you obtain equal to , then substitute in x=1.

    Spoiler:
    Show
    First one would be \frac{dy}{dx}=8x, so follow on from there.
    for each curve find the gradient at the given point.

    y=4x^2+2 at x=1 gradient= 8

    y= x^3-2 at x=1 gradient= 3

    y= 4x^3 at x=1 gradient=12

    are they right
  16. Offline

    ReputationRep:
    (Original post by jetskiwavedunkno)
    for each curve find the gradient at the given point.

    y=4x^2+2 at x=1 gradient= 8

    y= x^3-2 at x=1 gradient= 3

    y= 4x^3 at x=1 gradient=12

    are they right
    yes they are all right.
  17. Offline

    ReputationRep:
    (Original post by jetskiwavedunkno)
    for each curve find the gradient at the given point.

    y=4x^2+2 at x=1 gradient= 8

    y= x^3-2 at x=1 gradient= 3

    y= 4x^3 at x=1 gradient=12

    are they right
    Yes, they are.
  18. Offline

    ReputationRep:
    (Original post by jetskiwavedunkno)
    you are wrong


    (Original post by jordan-s)
    I'm afraid the gradient at 3 is also incorrect?

    3x^2 when x = 3 gives 3 multiplied by 3^2, which is 3 x 9 = 27..

    Ooops, too late! You've got it now anyway.
    Sorry, i also calculated it as 27 but i don't know why did i wrote the gradient at x=3 is correct. Actually i was busy at another thread as well so i just got confused.

    Btw, thanks for indicating my mistake.
  19. Offline

    ReputationRep:
    (Original post by raheem94)
    Sorry, i also calculated it as 27 but i don't know why did i wrote the gradient at x=3 is correct. Actually i was busy at another thread as well so i just got confused.

    Btw, thanks for indicating my mistake.
    dw about it raheem
  20. Offline

    ReputationRep:
    (Original post by Mr Inquisitive)
    Yes, they are.
    Thank you sir

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide to join now Processing…

Updated: February 23, 2012
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

New on TSR

Rate your uni

Help build a new league table

Poll
How do you read?
Study resources
Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.