Question

Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
2(x^2+y^2)^2=25(x^2−y^2)at point (3,1)

Answer 1

\[2(x^2+y^2)2=25(x^2-y^2)\]Is this what you mean?
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Answer 2

no i fixed it the first part is squared

Answer 3

\[2(x^2+y^2)^2=25(x^2-y^2)\]Is it like this then?
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Answer 4

yeah thats it

Answer 5

So let's first expand and separate the y's and x's and then we can implicitly differentiate both sides just to make things easier.\[2(x^2+y^2)^2=25(x^2-y^2) \rightarrow 4(x^2+y^2)(2x+2y \frac{ dy }{ dx })=25(2x-2y \frac{ dy }{ dx })\]\[(x^2+y^2)(x+y \frac{ dy }{ dx })=25/4(x-y \frac{ dy }{ dx }) \rightarrow x^3+x^2y \frac{ dy }{ dx }+xy^2+y^3\frac{ dy }{ dx }=\frac{ 25 }{ 4 }(x-y \frac{ dy }{ dx })\]\[\rightarrow x^3+x^2y \frac{ dy }{ dx }+xy^2+y^3\frac{ dy }{ dx }=\frac{ 25 }{ 4 }x-\frac{ 25 }{ 4 }y\frac { dy }{ dx } \]\[\rightarrow x^2y\frac { dy }{ dx }+y^3\frac { dy }{ dx }+\frac{ 25 }{ 4 }y\frac { dy }{ dx } =\frac{ 25 }{ 4 }x-x^3-xy^2 \rightarrow \frac { dy }{ dx }(x^2y+y^3+\frac{ 25 }{ 4 }y)=\frac{ 25 }{ 4 }x-x^3-xy^2 \]\[\rightarrow \frac{ dy }{ dx }=\frac{ 25x-4x^3-4xy^2 }{ 25y+4x^2y+4y^3 }\]For point (3, 1), we plug in 3 and 1 for x and y respectively and evaluate dy/dx. \[\frac{ dy }{ dx }=\frac{ 25(3)-4(3)^3-4(3)(1)^2 }{ 25(1)+4(3)^2(1)+4(1)^3 }=\frac{ 75-108-12 }{25+36+4 }=-\frac{ 45 }{ 65 }=-\frac{ 9 }{ 11 }\]We know the slope now, just figure out the equation by plugging in the given point and that's it.

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Answer 6

The thing about implicit differentiation is to first differentiate both sides with respect to x, then expand and separate all terms containing dy/dx on one side and all the other terms on the other side. Now factor out dy/dx from all the terms containing dy/dx on the isolated side and that way you can solve for dy/dx. That's the key to implicit differentiation.

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