Join TSR now and get all your revision questions answeredSign up now

need help on dy/dx Watch

    • Thread Starter
    Offline

    0
    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??
    Offline

    0
    ReputationRep:
    (Original post by jetskiwavedunkno)
    is it x=3 graient =27
    x=5 gradient =75

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

    0
    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
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Gemini92)
    Those are all right.
    awww thank you! xx
    Offline

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

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

    0
    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.
    • Thread Starter
    Offline

    0
    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
    Offline

    0
    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)
    • Community Assistant
    • Welcome Squad
    Offline

    3
    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
    • Thread Starter
    Offline

    0
    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??
    • Thread Starter
    Offline

    0
    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??
    • Community Assistant
    • Welcome Squad
    Offline

    3
    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.
    • Thread Starter
    Offline

    0
    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
    Offline

    3
    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.
    • Thread Starter
    Offline

    0
    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
    • Community Assistant
    • Welcome Squad
    Offline

    3
    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.
    Offline

    3
    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.
    Offline

    1
    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.
    • Thread Starter
    Offline

    0
    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
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Mr Inquisitive)
    Yes, they are.
    Thank you sir
 
 
 
Poll
Is GoT overrated?
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

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.