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

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

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

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

nickymardenBest ResponseYou've already chosen the best response.0
yeah, i did that :)
 one year ago

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

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

myininayaBest 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
 one year ago

robtobeyBest 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)\[\frac{\left(3\ 1^2+1^2\right)}{2 (12) 1}=2 \]
 one year ago
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