SOAL $b^{2}x^{2}+x^{2}y^{2}=a^{2}y^{2}$ $b^{2}x^2=a^{2}y^2-x^{2}y^2$ $b^{2}x^2=y^{2}(a^2-x^2)$ $y^2=\frac{b^{2}x^2}{a^2-x^2}$ $y=\sqrt{\frac{b^{2}x^2}{a^2-x^2}}$ $y=\frac{bx}{\sqrt{a^2-x^2}}$ Missalkan: $u(x)=bx\to u’(x)=b$ $v(x)=\sqrt{a^2-x^2}\to v’(x)=\frac{1}{2}(a^2-x^2)^{\frac{-1}{2}}(2x)=\frac{-x}{\sqrt{a^2-x^2}}$ $y=\frac{u(x)}{v(x)}\to y’=\frac{u’(x).v(x)-v’(x).u(x)}{v^{2(x)}}$ $y’=\frac{b.\sqrt{a^2-x^2}-\frac{-x.bx}{\sqrt{a^2-x^2}}}{(\sqrt{a^2-x^2})^2}$ $y’=\frac{b.(\sqrt{a^2-x^2})^2+bx^2}{\sqrt{a^2-x^2}.(a^2-x^2)}$ $y’=\frac{b.(a^2-x^2)+bx^2}{\sqrt{a^2-x^2}.(a^2-x^2)}$ $y’=\frac{b.a^2-bx^2+bx^2}{\sqrt{a^2-x^2}.(a^2-x^2)}$ $y’=\frac{b.a^2}{(a^2-x^2)^{\frac{3}{2}}}$ Nilai maks/min $y’=0$ $0=\frac{b.a^2}{(a^2-x^2)^{\frac{3}{2}}}$ $ba^2=0$ Jadi fungsinya tidak memiliki nilai maksimum/minimum. Titik Belok; $y’’=0$ $y’=\frac{b.a^2}{(a^2-x^2)^{\frac{3}{2}}}$ Misalkan: $p(x)=
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