Question Number 227073 by Spillover last updated on 29/Dec/25
Answered by Kassista last updated on 29/Dec/25
$$ \\ $$$$\frac{{dy}}{{dx}}=\:\frac{−\left({x}−{b}\right)+\left({a}−{x}\right)}{\mathrm{2}\sqrt{\left({a}−{x}\right)\left({x}−{b}\right)}}−\left({a}−{b}\right).\frac{\mathrm{1}}{\mathrm{1}+\left(\sqrt{\frac{{a}−{x}}{{x}−{b}}}\right)^{\mathrm{2}} }.\frac{\frac{−\left({x}−{b}\right)−\left({a}−{x}\right)}{\left({x}−{b}\right)^{\mathrm{2}} }}{\mathrm{2}\sqrt{\frac{{a}−{x}}{{x}−{b}}}} \\ $$$$ \\ $$$$ \\ $$$$\therefore\frac{{dy}}{{dx}}=\frac{{b}+{a}−\mathrm{2}{x}}{\mathrm{2}\sqrt{\left({a}−{x}\right)\left({x}−{b}\right)}}−\left({a}−{b}\right).\frac{\mathrm{1}}{\frac{{x}−{b}+{a}−{x}}{{x}−{b}}}.\frac{{b}−{a}}{\left({x}−{b}\right)^{\mathrm{2}} }.\frac{\sqrt{{x}−{b}}}{\mathrm{2}\sqrt{{a}−{x}}} \\ $$$$ \\ $$$$\therefore\frac{{dy}}{{dx}}=\frac{{b}+{a}−\mathrm{2}{x}}{\mathrm{2}\sqrt{\left({a}−{x}\right)}\sqrt{\left({x}−{b}\right)}}−\left({a}−{b}\right).\frac{{x}−{b}^{\mathrm{1}} }{{a}−{b}}.\frac{{b}−{a}}{\left({x}−{b}\right)^{\mathrm{2}} }.\frac{\left({x}−{b}\right)^{\mathrm{0}.\mathrm{5}} }{\mathrm{2}\sqrt{{a}−{x}}} \\ $$$$ \\ $$$$ \\ $$$$\therefore\frac{{dy}}{{dx}}=\frac{{b}+{a}−\mathrm{2}{x}}{\mathrm{2}\sqrt{{a}−{x}}\sqrt{{x}−{b}}}−\frac{{b}−{a}}{\:\sqrt{{x}−{b}}}.\frac{\mathrm{1}}{\mathrm{2}\sqrt{{a}−{x}}} \\ $$$$ \\ $$$$\therefore\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{{a}−{x}}\sqrt{{x}−{b}}}\left(\mathrm{2}{a}−\mathrm{2}{x}\right)=\frac{\left({a}−{x}\right)}{\:\sqrt{{a}−{x}}\sqrt{{x}−{b}}} \\ $$$$ \\ $$$$\therefore\frac{{dy}}{{dx}}=\frac{\sqrt{{a}−{x}}}{\:\sqrt{{x}−{b}}}=\sqrt{\frac{{a}−{x}}{{x}−{b}}} \\ $$$$ \\ $$