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26 the ti•curve y , where to curve has

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$$\begin{gathered} Given : A Curve y=\frac{x}{1+x^2} \\ To Calculate : Points on the curve where \tan gent to the curve has greatest slope. \\ solution : given curve is y=\frac{x}{1+x^2} \\ slope of \tan gent at any point of the curve is given by \frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx}=\frac{\left\{\right(1+x^2\left){\displaystyle \frac{d}{dx}}\right(x\left)\right\}-\left\{\right(x\left){\displaystyle \frac{d}{dx}}\right(1+x^2\left)\right\}}{(1+x^2{)}^2} \\ \Rightarrow \frac{dy}{dx}=\frac{\left\{\right(1+x^2\left)\right(1\left)\right\}-\left\{\right(x\left)\right(0+2x\left)\right\}}{(1+x^2{)}^2} \\ \Rightarrow \frac{dy}{dx}=\frac{(1+x^2-2x^2)}{(1+x^2{)}^2}=\frac{1-x^2}{(1+x^2{)}^2} \\ and for greatest slope\hspace{0.17em}(Slope maximum), we need to maximise \frac{dy}{dx} \\ so, assuming \frac{dy}{dx}=k \\ \Rightarrow k=\frac{1-x^2}{(1+x^2{)}^2} \\ to maximise k, puttting \frac{dk}{dx}=0 \\ \Rightarrow \frac{dk}{dx}=\frac{\left\{\right(1+x^2{)}^2{\displaystyle \frac{d}{dx}}(1-x^2)\}-\{(1-x^2){\displaystyle \frac{d}{dx}}(1+x^2{)}^2\}}{(1+x^2{)}^4}=0 \\ \Rightarrow \left\{\right(1+x^2{)}^2(0-2x)\}-\{(1-x^2)2(1+x^2)(0+2x)\}=0\times (1+x^2{)}^4 \\ \Rightarrow -2x(1+x^2{)}^2-4x(1-x^2\left)\right(1+x^2)=0 \\ \Rightarrow -2x(1+x^2\left)\right\{(1+x^2)+2(1-x^2)\}=0 \\ \Rightarrow -2x(1+x^2\left)\right(1+x^2+2-2x^2)=0 \\ \Rightarrow -2x(1+x^2\left)\right(3-x^2)=0 \\ \Rightarrow -2x(1+x^2\left)\right(-x^2+3)=0 \\ \Rightarrow either -2x=0 \Rightarrow x=0 \\ or 1+x^2=0 \Rightarrow x^2=-1 (Which is not a Real value) \\ or -x^2+3=0\Rightarrow x^2=3\Rightarrow x=\pm \sqrt{3} \\ So, x can take 0, \pm \sqrt{3} \\ putting all these values in k=\frac{1-x^2}{(1+x^2{)}^2} and check which gives maximum value \\ for x=0 \\ k=\frac{1-0^2}{(1+0^2{)}^2}=1 \\ for x=\pm \sqrt{3} \\ k=\frac{1-(\pm \sqrt{3}{)}^2}{(1+(\pm \sqrt{3}{)}^2{)}^2}=\frac{1-{\displaystyle 3}}{(1+3{)}^2}=\frac{{\displaystyle -2}}{{\displaystyle 16}}=\frac{-1}{8} \\ so, at x=0, k=1 is a greater value than at x=\pm \sqrt{3}, k=\frac{-1}{8} \\ so, for x=0, Curve y=\frac{x}{1+x^2} will have greatest slope \\ and at x=0, y=0 (putting x=0 in Curve y=\frac{x}{1+x^2}) \\ Hence, point at which \tan gent to the curve has greatest slope is (0,0). \end{gathered}$$
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