Why is the Coordinate of Point P (3/2, 0) in this Differentiation Problem?

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The discussion centers on finding the coordinates of point P (3/2, 0) related to the curve defined by the equation x²y = x - 6. The original method involved differentiating the equation after rearranging it to find y, resulting in an incorrect derivative of f'(x) = 1 + 12x⁻³. The correct derivative, derived from the rearranged equation y = (x - 6)/x², is (12 - x)/x³. This correction leads to the accurate identification of point P.

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"the equation of a curve is x^2y=x-6 (x^2y is x squared times y)

The normal of the curve at the point where x=-2 meets the x-axis at p.

Find the coordinates of P"


The method I used was to find y by rearranging the given equation and differentiate it.
This gave me f ' (x)= 1+12x^-3

Then I subbed x=-2 in so that I could find the gradient of the tangent at that point. It worked out as -1/2, so then the gradient of the normal would be 2.

After finding the equation of the line and making y=0, I got the wrong answer: x=-1 and y=0

apparently the answer is (3/2, 0)


any help would be appreciated
 
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your derivative is wrong
get it right , you will have your answer.
 
If you rearrange the equation to y = (x - 6)/x^2, the derivative is (12-x)/x^3.
 

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