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need help on dy/dx

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Applying to Uni? Let Universities come to you. Click here to get your perfect place 20-10-2014
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    Find the equations of the normals to the curves at the given point

    y=(x+1)^2 at x=0

    y=4/x at x=1

    answers in form ax+by+c=0

    how do i do this

    also can some1 check if this is right

    find the equations of the tangents to the curves at the given points

    y=5x^2+1 at x=2 y=50?

    y=3x^2-4x at x=3 y=14??
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    For the first part is not right, apart from x=2. To calculated the gradient of the tangent, you differentiate x^2, and then substitute your values for x to get the gradient.

    Similar thing for x^3 except you differentiate x^3 instead.
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    (Original post by mr tim)
    For the first part is not right, apart from x=2. To calculated the gradient of the tangent, you differentiate x^2, and then substitute your values for x to get the gradient.

    Similar thing for x^3 except you differentiate x^3 instead.
    how do i do it then
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    (Original post by mr tim)
    For the first part is not right, apart from x=2. To calculated the gradient of the tangent, you differentiate x^2, and then substitute your values for x to get the gradient.

    Similar thing for x^3 except you differentiate x^3 instead.
    ah wait for the first ones is it

    x=2 gradient=4

    x=6 gradient=12

    x=-1 gradient=1?
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    (Original post by jetskiwavedunkno)
    ah wait for the first ones is it

    x=2 gradient=4

    x=6 gradient=12

    x=-1 gradient=1?
    The first two are right, but the last one is wrong.
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    (Original post by jetskiwavedunkno)
    ah wait for the first ones is it

    x=2 gradient=4

    x=6 gradient=12

    x=-1 gradient=1?
    Your calculated gradient at x=-1 is wrong, others are correct.
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    (Original post by Gemini92)
    The first two are right, but the last one is wrong.
    x=-1 gradient= -2??? how do i do the x^3 ones
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    (Original post by raheem94)
    Your calculated gradient at x=-1 is wrong, others are correct.
    is the gradient -2? nd how do i do the x^3 ones plz
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    (Original post by jetskiwavedunkno)
    is the gradient -2? nd how do i do the x^3 ones plz
    Yes the gradient is -2.

    For the x^3 expression, first differentiate it, than substitute different values of x to find the gradient at different points.
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    (Original post by jetskiwavedunkno)
    is the gradient -2? nd how do i do the x^3 ones plz
    To differentiate, you'll multiply x by its power, and then subtract 1 from the power.

    Eg. differentiating x^6 =6x^5.

    Yeah, -2's correct. 2(-1)=-2.

    Try using the rule I stated above to differentiate x^3. Actually write what you get in terms of x first so we can see what you're doing.
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    (Original post by jetskiwavedunkno)
    x=-1 gradient= -2??? how do i do the x^3 ones
    That's right. For the x^3 ones, you have to differentiate x^3 and then substitute each x value into the derivative to get the gradient.
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    For the first one, you need to start by differentiating x^2.
    Differentiating it gives you "dy/dy", also known as the gradient function.
    The method of differentiating, is "Multiply by the current power, and then decrease the power by 1."

    So to differentiate x^2, you multiply by 2, and decrease the power fromk 2 to 1; hence the result is 2x.

    Youy would then need to substitute the value of x in to find the gradient at that point.

    For x^3, it is a similar process. Start by multiplying by the power, so multiply by 3, and then reduce the power from 3 to 2. Hence the result is 3x^2. Again sub in values of x to find the gradient.
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    (Original post by raheem94)
    Yes the gradient is -2.

    For the x^3 expression, first differentiate it, than substitute different values of x to find the gradient at different points.
    is it for x^3

    x=3 gradient=27

    x=5 gradient=75

    x=-3 gradient=27

    ????
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    (Original post by Contrad!ction.)
    To differentiate, you'll multiply x by its power, and then subtract 1 from the power.

    Eg. differentiating x^6 =6x^5.

    Yeah, -2's correct. 2(-1)=-2.

    Try using the rule I stated above to differentiate x^3. Actually write what you get in terms of x first so we can see what you're doing.
    is it for x^3

    x=3 gradient=27

    x=5 gradient=75

    x=-3 gradient=27

    ????
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    (Original post by Gemini92)
    That's right. For the x^3 ones, you have to differentiate x^3 and then substitute each x value into the derivative to get the gradient.
    is it for x^3

    x=3 gradient=27

    x=5 gradient=75

    x=-3 gradient=27

    ????
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    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.
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    (Original post by jordan-s)
    For the first one, you need to start by differentiating x^2.
    Differentiating it gives you "dy/dy", also known as the gradient function.
    The method of differentiating, is "Multiply by the current power, and then decrease the power by 1."

    So to differentiate x^2, you multiply by 2, and decrease the power fromk 2 to 1; hence the result is 2x.

    Youy would then need to substitute the value of x in to find the gradient at that point.

    For x^3, it is a similar process. Start by multiplying by the power, so multiply by 3, and then reduce the power from 3 to 2. Hence the result is 3x^2. Again sub in values of x to find the gradient.
    ahhh crap wait so differentiating 3x is 3x^2 and not 2x^2??
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    No.. differentiating x^3 gives 3x^2.. sub in x = 3 and you get 3(3)^2 = 27.
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    (Original post by jetskiwavedunkno)
    is it for x^3

    x=3 gradient=18

    x=5 gradient=50

    x=-3 gradient=18

    ????
    Try again, all are wrong.

    Differentiating x^3 gives 3x^2.
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    (Original post by jetskiwavedunkno)
    ahhh crap wait so differentiating 3x is 3x^2 and not 2x^2??
    Yes, the differential of x^3 is 3x^2

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Updated: February 23, 2012
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