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Diffentiation - C3 Q2 june 2010 help (edexcel) Watch

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    I've got up to -6(5-3x)^-3 = 3, Can I get some help finishing this question? thanks.
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    (Original post by ZombieCombat)

    I've got up to -6(5-3x)^-3 = 3, Can I get some help finishing this question? thanks.
    You've differentiated incorrectly.

    Once you find dy/dx in terms of x, sub x=2 into it to get a value for dy/dx. Remember that the normal is perpendicular to the tangent (dy/dx) of the curve at x =2 thus find the value of dy/dx of the normal. Plug in your values into the equation y-y_1=m(x-x_1) where P(x_1,y_1) and m is the gradient of the normal then simplify.
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    (Original post by ZombieCombat)

    I've got up to -6(5-3x)^-3 = 3, Can I get some help finishing this question? thanks.
    your equation is not right
    The left side is the incomplete derivatice of C, there sould be a factor
    of -3 over there.
    ísubstitutingthe given x0 in it you will get the m value of tangent at P.
    The direction factor of the normal will be -1/m.
    At the right side of your equation there is the y coordinate of P (y0)
    the equation of normal y-y0=-1/m(x-x0)
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    Complete the derivative of c, then dy/dx using chain rule, finding perpendicular gradient hence giving gradient of normal.
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    Ok thanks guys got the right answer =D.
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    Heey !
    Does any one have june 2010 - edexcel C3 paper?
    Please if you do, send me the link
    thanks xxx
 
 
 
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