need help on dy/dx

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

    x=-3 gradient =27??
    Those are all right.
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    (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
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    (Original post by Gemini92)
    Those are all right.
    awww thank you! xx
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    (Original post by jetskiwavedunkno)
    is it x=3 graient =27
    x=5 gradient =75

    x=-3 gradient =27??
    Yes, they're correct.
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    (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.
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    (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
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    (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)
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    (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
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    (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??
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    (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??
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    (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.
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    (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
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    (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.
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    (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
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    (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.
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    (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.
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    (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.
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    (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
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    (Original post by Mr Inquisitive)
    Yes, they are.
    Thank you sir
 
 
 
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