Question

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 ).

Answer 1

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

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

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