+26400 sqrt(1-x^2)-x/2 What are the 0 points of the derivative function?
\(\begin{array}{rcll} y = \sqrt{1-x^2}- \frac{x}{2} &=& (1-x^2)^{\frac12} -\frac{x}{2}\\\\ y' &=& \frac12 \cdot (1-x^2)^{\frac12 -1} \cdot (-2x) - \frac12 \\ y' &=& \frac12 \cdot (1-x^2)^{-\frac12} \cdot (-2x) - \frac12 \\ y' &=& (1-x^2)^{-\frac12} \cdot (-x) - \frac12 \\ y' &=& -\frac{x}{ \sqrt{1-x^2} }- \frac12 \\ y'=0\\ -\frac{x}{ \sqrt{1-x^2} }- \frac12 &=& 0 \\ -\frac{x}{ \sqrt{1-x^2} } &=& \frac12 \\ -2x &=& \sqrt{1-x^2} \qquad | \qquad ()^2\\ (-2x)^2 &=& (\sqrt{1-x^2})^2 \\ 4x^2 &=& 1-x^2\\ 5x^2 &=& 1\\ x^2 &=& \frac15 \qquad | \qquad \sqrt{} \\ x_{1,2} &=& \pm \sqrt{\frac15} \\\\ x_1 &=& \sqrt{\frac15}\\ x_1 &=& \frac{1}{\sqrt{5} } \qquad \text{ no solution!}\\\\ x_2 &=& -\sqrt{\frac15} \\ x_2 &=& -\frac{1}{\sqrt{5} } \end{array}\)
\(\begin{array}{rcll} \text{The solution is } \quad x = -\frac{1}{\sqrt{5} } \end{array}\)
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+26400 sqrt(1-x^2)-x/2 What are the 0 points of the derivative function?
\(\begin{array}{rcll} y = \sqrt{1-x^2}- \frac{x}{2} &=& (1-x^2)^{\frac12} -\frac{x}{2}\\\\ y' &=& \frac12 \cdot (1-x^2)^{\frac12 -1} \cdot (-2x) - \frac12 \\ y' &=& \frac12 \cdot (1-x^2)^{-\frac12} \cdot (-2x) - \frac12 \\ y' &=& (1-x^2)^{-\frac12} \cdot (-x) - \frac12 \\ y' &=& -\frac{x}{ \sqrt{1-x^2} }- \frac12 \\ y'=0\\ -\frac{x}{ \sqrt{1-x^2} }- \frac12 &=& 0 \\ -\frac{x}{ \sqrt{1-x^2} } &=& \frac12 \\ -2x &=& \sqrt{1-x^2} \qquad | \qquad ()^2\\ (-2x)^2 &=& (\sqrt{1-x^2})^2 \\ 4x^2 &=& 1-x^2\\ 5x^2 &=& 1\\ x^2 &=& \frac15 \qquad | \qquad \sqrt{} \\ x_{1,2} &=& \pm \sqrt{\frac15} \\\\ x_1 &=& \sqrt{\frac15}\\ x_1 &=& \frac{1}{\sqrt{5} } \qquad \text{ no solution!}\\\\ x_2 &=& -\sqrt{\frac15} \\ x_2 &=& -\frac{1}{\sqrt{5} } \end{array}\)
\(\begin{array}{rcll} \text{The solution is } \quad x = -\frac{1}{\sqrt{5} } \end{array}\)
