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Question Number 35642 by rahul 19 last updated on 21/May/18

$$Ify=\sqrt{(a-x)(x-b)}-(a-b){\tan }^{-1}\left({\left(\frac{a-x}{x-b}\right)}^{0.5}\right).$$$$Thenfind\frac{dy}{dx}?$$

Commented by rahul 19 last updated on 21/May/18

$$Ididbyusingthesubstitution$$$$x=a\cos {}^2\theta +b\sin {}^2\theta .$$$$But{i}^{\prime }mgettingwronganswer.$$

Commented by abdo imad last updated on 21/May/18

$$wehavey\left(x\right)=\sqrt{(a-x)(x-b)}-(a-b)arctan\left(\sqrt{\frac{a-x}{x-b}}\right)$$$$y\left(x\right)=\sqrt{u\left(x\right)}-(a-b)arctan\left(v\left(x\right)\right)\Rightarrow$$$$y^{{}^{\prime }}\left(x\right)=\frac{u^{{}^{\prime }}\left(x\right)}{2\sqrt{u\left(x\right)}}-(a-b)\frac{v^{{}^{\prime }}\left(x\right)}{1+v^2\left(x\right)}butwehave$$$$u^{{}^{\prime }}\left(x\right)=-(x-b)+a-x=-2x+a+b{\mid }_1^{-1}$$$$v^{{}^{\prime }}\left(x\right)=\frac{{\left(\frac{a-x}{x-b}\right)}^{{}^{\prime }}}{2\sqrt{\frac{a-x}{x-b}}}=\frac{-(x-b)-(a-x)}{{(x-b)}^{2}2.\sqrt{\frac{a-x}{x-b}}}=\frac{b-a}{2{(x-b)}^2\frac{\sqrt{a-x_{}}}{\sqrt{x-b}}}$$$$=\frac{b-a}{2{\left(\sqrt{x-b}\right)}^3\sqrt{a-x}}\Rightarrow$$$$y^{{}^{\prime }}\left(x\right)=\frac{-2x+a+b}{2\sqrt{(a-x)(x-b)}}+\frac{{(b-a)}^2}{2\sqrt{a-x}{\left(\sqrt{x-b}\right)}^3}\times \frac{1}{1+\frac{a-x}{x-b}}$$$$=\frac{-2x+a+b}{2\sqrt{(a-x)(x-b)}}+\frac{{(b-a)}^2}{2}\frac{x-b}{2\sqrt{a-x}(x-b)\sqrt{x-b}(a-b)}$$$$=\frac{-2ax+a+b}{2\sqrt{(a-x)(x-b)}}+\frac{a-b}{4\sqrt{a-x}\sqrt{x-b}}.$$

Answered by ajfour last updated on 21/May/18

$$\frac{dy}{dx}=\frac{-2x+(a+b)}{2\sqrt{(a-x)(x-b)}}-\frac{(a-b)}{1+\frac{a-x}{x-b}}\times \frac{\frac{-\sqrt{x-b}}{2\sqrt{a-x}}-\frac{\sqrt{a-x}}{2\sqrt{x-b}}}{x-b}$$$$=\frac{-2x+(a+b)}{2\sqrt{(a-x)(x-b)}}-\frac{b-a}{2\sqrt{(a-x)(x-b)}}$$$$\frac{dy}{dx}=\sqrt{\frac{a-x}{x-b}}.$$

Commented by rahul 19 last updated on 21/May/18

$$Thankyousir.$$

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