Answer to Question #247495 in Calculus for Jaymz

Question #247495
Use Implicit differentiation to find the slope of the tangent line of the curve (x^2+y^2)^5 = ((x^2-y^2)^2 - 4x^2y^2)^2 at the point ( √2/2 , √2/2 ).
1
Expert's answer
2022-02-07T16:43:12-0500

"\\text{The expression given is:} \\to (x^2+y^2)^5 = ((x^2-y^2)^2 - 4x^2y^2)^2\n\\\\ \\text{We differentiate:} \\to (x^2+y^2)^{5\/2} = (x^2-y^2)^2 - 4x^2y^2\n\\\\ \\text{As a result from the latter we have:} \n\\\\ 5(x^2+y^2)^{3\/2} (x+y\\cdot y') = 4(x^2-y^2)(x-y\\cdot y') - 8xy(y+x\\cdot y')"


"\\\\ \\text{We substitute the point P (\u221a2\/2 , \u221a2\/2) to find y}':\n\\\\ 5((\\frac{\\sqrt{2}}{2})^2+(\\frac{\\sqrt{2}}{2})^2)^{3\/2} (\\frac{\\sqrt{2}}{2}+(\\frac{\\sqrt{2}}{2})\\cdot y') = 4((\\frac{\\sqrt{2}}{2})^2-(\\frac{\\sqrt{2}}{2})^2)(\\frac{\\sqrt{2}}{2}-(\\frac{\\sqrt{2}}{2})\\cdot y') - 8(\\frac{\\sqrt{2}}{2})(\\frac{\\sqrt{2}}{2})(\\frac{\\sqrt{2}}{2}+(\\frac{\\sqrt{2}}{2})\\cdot y')\n\\\\ \\to 5(\\frac{1}{2}+\\frac{1}{2})^{3\/2}(\\frac{\\sqrt{2}}{2})\\cdot (1+ y') = 4(\\frac{1}{2}-\\frac{1}{2})(\\frac{\\sqrt{2}}{2})\\cdot(1- y') - 8(\\frac{1}{2})(\\frac{\\sqrt{2}}{2})\\cdot (1+y')\n\\\\ \\to 5(1)^{3\/2}\\cancel{(\\frac{\\sqrt{2}}{2})}\\cdot (1+ y') =- 8(\\frac{1}{2})\\cancel{(\\frac{\\sqrt{2}}{2})}\\cdot (1+y')\n\\\\ \\to 5\\cdot (1+ y') =- 4\\cdot (1+y')\n\\\\ \\to 5+5\\cdot y' =- 4- 4\\cdot y' \n\\\\ \\to 5+4 = 9 =-5\\cdot y' - 4\\cdot y' = -9\\cdot y' \\to y' = \\frac{9}{-9}=-1"


In conclusion, the slope of the curve tangent to the line at the point P(½,½) is y'= -1.


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