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

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

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

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

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

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

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

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

nickymarden Group TitleBest ResponseYou've already chosen the best response.0
thank you :))
 2 years ago

robtobey Group TitleBest 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)\[\frac{\left(3\ 1^2+1^2\right)}{2 (12) 1}=2 \]
 2 years ago
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