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# Diffentiation - C3 Q2 june 2010 help (edexcel) Watch

1. I've got up to -6(5-3x)^-3 = 3, Can I get some help finishing this question? thanks.
2. (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.
3. (Original post by ZombieCombat)

I've got up to -6(5-3x)^-3 = 3, Can I get some help finishing this question? thanks.
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)
4. Complete the derivative of c, then dy/dx using chain rule, finding perpendicular gradient hence giving gradient of normal.
5. Ok thanks guys got the right answer =D.
6. Heey !
Does any one have june 2010 - edexcel C3 paper?
thanks xxx

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