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nickymarden
 3 years ago
Determine the tangent line to he curve y^2(2x)=x^3 at the point P(1,1).
nickymarden
 3 years ago
Determine the tangent line to he curve y^2(2x)=x^3 at the point P(1,1).

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myko
 3 years ago
Best ResponseYou've already chosen the best response.0put it like this: \[y=\pm \sqrt{x ^{3}/(2x)}\] abd find the derivative

nickymarden
 3 years ago
Best ResponseYou've already chosen the best response.0i did \[f(1) =1\] \[(f(x))^2.(2x)=x^3\] and derived it

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.2\[y^2(2x)=x^3\] \[(y^2)'(2x)+y^2(2x)'=(x^3)' \text{ product rule }\] \[2yy'(2x)+y^2(01)=3x^2\]

nickymarden
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, i did that :)

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.2\[2yy'(2x)+y^2(1)=3x^2\] \[2yy'(2x)y^2=3x^2\]

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.2\[\text{ add } y^2 \text{ on both sides }\] \[\text{ divide both sides by } 2y(2x) \]

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.2The objective is to solve for y' Following these steps will do that y' is the slope by the way

robtobey
 3 years ago
Best ResponseYou've already chosen the best response.0\[(2x) y^2=x^3 \]Calculate the total derivative of the above\[2 (2x) y dyy^2 dx=3x^2 dx \]Solve for dy\[dy=\frac{\left(3 x^2+y^2\right) dx}{2 (x2) y} \]then divide both sides by dx\[\frac{dy}{ dx}=\frac{\left(3 x^2+y^2\right)}{2 (x2) y} \]Evaluate the RHS of the above at point (1,1)\[\frac{\left(3\ 1^2+1^2\right)}{2 (12) 1}=2 \]
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