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nickymarden
Determine the tangent line to he curve y^2(2-x)=x^3 at the point P(1,1).
put it like this: \[y=\pm \sqrt{x ^{3}/(2-x)}\] abd find the derivative
i did \[f(1) =1\] \[(f(x))^2.(2-x)=x^3\] and derived it
\[y^2(2-x)=x^3\] \[(y^2)'(2-x)+y^2(2-x)'=(x^3)' \text{ product rule }\] \[2yy'(2-x)+y^2(0-1)=3x^2\]
yeah, i did that :)
\[2yy'(2-x)+y^2(-1)=3x^2\] \[2yy'(2-x)-y^2=3x^2\]
\[\text{ add } y^2 \text{ on both sides }\] \[\text{ divide both sides by } 2y(2-x) \]
The objective is to solve for y' Following these steps will do that y' is the slope by the way
\[(2-x) y^2=x^3 \]Calculate the total derivative of the above\[2 (2-x) y dy-y^2 dx=3x^2 dx \]Solve for dy\[dy=-\frac{\left(3 x^2+y^2\right) dx}{2 (x-2) y} \]then divide both sides by dx\[\frac{dy}{ dx}=-\frac{\left(3 x^2+y^2\right)}{2 (x-2) y} \]Evaluate the RHS of the above at point (1,1)\[-\frac{\left(3\ 1^2+1^2\right)}{2 (1-2) 1}=2 \]