Differentiation (Part-38) | S N De

 $~7(i)~~y=x^3\cdot \sqrt{\frac{x^2+4}{x^2+3}}$

Sol. $~~~y=x^3\cdot \sqrt{\frac{x^2+4}{x^2+3}}\\ \text{or,}~~   \log y=\log\left(x^3\cdot \sqrt{\frac{x^2+4}{x^2+3}}\right) \\ \text{or,}~~  \log y= \log x^3+ \log \left(\sqrt{\frac{x^2+4}{x^2+3}}\right)\\ \text{or,}~~\log y=3\log x+\log\left(\frac{x^2+4}{x^2+3}\right)^{1/2} \\ \text{or,}~~ \log y= 3\log x+ \frac 12 \log\left(\frac{x^2+4}{x^2+3}\right) \\ \text{Differentiating w.r.t.}~~x, \\~~~~ \frac 1y ~\frac{dy}{dx}\\=\frac 3x+ \frac 12\cdot \frac{1}{\frac{x^2+4}{x^2+3}}\cdot \frac{d}{dx} \left(\frac{x^2+4}{x^2+3}\right)\\=\frac 3x+\frac 12 \cdot \frac{x^2+3}{x^2+4}\cdot \frac{(x^2+3) \cdot 2x-(x^2+4) \cdot 2x}{(x^2+3)^2}\\=\frac 3x+\frac 12 \cdot \frac{1}{x^2+4} \cdot \frac{2x(x^2+3-x^2-4)}{x^2+3}\\=\frac 3x-\frac{x}{(x^2+4)(x^2+3)}\\=\frac{3(x^2+4)(x^2+3)-x^2}{x(x^2+3)(x^2+4)} \\=\frac{3(x^4+7x+12)-x^2}{x(x^2+3)(x^2+4)}\\=\frac{3x^4+20x+36}{x(x^2+3)(x^2+4)} \\ \therefore \frac{dy}{dx}\\=y \cdot \frac{3x^4+20x+36}{x(x^2+3)(x^2+4)}\\= x^3\cdot \sqrt{\frac{x^2+4}{x^2+3}} \cdot\frac{3x^4+20x+36}{x(x^2+3)(x^2+4)} \\=\frac{x^2(3x^4+20x^2+36)}{(x^2+3)\sqrt{(x^2+3)(x^2+4)}}~~\text{(ans.)} $

$~~(ii)~~y=\sqrt{\frac{(x-a)(x-b)}{(x-c)(x-d)}}$

Sol. $~~~y=\sqrt{\frac{(x-a)(x-b)}{(x-c)(x-d)}}\\ \text{or,}~~  \log y=\log \left[\frac{(x-a)(x-b)}{(x-c)(x-d)}\right]^{1/2} \\ \text{or,}~~ \log y=\frac 12 [\log\{(x-a)(x-b)\}\\~~~~~~~~~~-\log\{(x-c)(x-d)\}] \\ \text{or,}~~ \log y=\frac 12 [\log(x-a)+\log(x-b)\\~~~~~~~~~~-\log(x-c)-\log(x-d)] \\ \text{ Diff. w.r.t.}~~x, \\~~\frac 1y~\frac{dy}{dx}\\=\frac 12[\frac{1}{x-a}+\frac{1}{x-b}-\frac{1}{x-c}-\frac{1}{x-d}] \\ \therefore~ \frac{dy}{dx}\\=\frac 12\sqrt{\frac{(x-a)(x-b)}{(x-c)(x-d)}} \left[\frac{1}{x-a}+\frac{1}{x-b}-\frac{1}{x-c}-\frac{1}{x-d}\right]~~\text{(ans.)}$

$~~(iii)~~y=\log\sqrt{\frac{1+\sin x}{1-\sin x}}$

Sol. $~~~~~~~~y=\log\sqrt{\frac{1+\sin x}{1-\sin x}}\\ \text{or,}~~ ~y=\log\left(\frac{1+\sin x}{1-\sin x}\right)^{1/2} \\ \text{or,}~~  y=\frac 12[\log(1+\sin x)-\log(1-\sin x)]\\ \therefore~~\frac{dy}{dx}\\=\frac 12 \left[\frac{1}{1+\sin x}\frac{d}{dx}(1+\sin x)\\~~~-\frac{1}{1-\sin x}\frac{d}{dx}(1-\sin x)\right]\\=\frac 12 \left[\frac{\cos x}{1+\sin x}-\frac{-\cos x}{1-\sin x}\right]\\=\frac 12\left[\frac{\cos x}{1+\sin x}+\frac{\cos x}{1-\sin x}\right]\\=\frac 12 \cos x \left[\frac{1}{1+\sin x}+\frac{1}{1-\sin x}\right]\\=\frac 12 \cos x \left[\frac{1-\sin x+1+\sin x}{(1+\sin x)(1-\sin x)}\right]\\=\frac 12 ~\cos x \cdot \frac{2}{1-\sin^2x}\\=\frac{\cos x}{\cos^2x}\\=\frac{1}{\cos x} \\ \therefore ~\frac{dy}{dx}=\sec x~~\text{(ans.)}$

$~~(iv)~~y=\frac{x+2}{(x-1)(x+5)}$

Sol. $~~~~~y=\frac{x+2}{(x-1)(x+5)} \\ \text{or,}~~ \log y=\log\left(\frac{x+2}{(x-1)(x+5)}\right)\\ \text{or,}~~ \log y= \log(x+2)\\~~~~~~~~~~~-\log[(x-1)(x+5)] \\ \text{or,}~~  \log y=\log(x+2)\\~~~~~~~~-[\log(x-1)+\log(x+5)] \\ \text{Diff. w.r.t.}~~x, \\ ~~~\frac 1y~\frac{dy}{dx}\\=\frac{1}{x+2}-\frac{1}{x-1}-\frac{1}{x+5}\\ \therefore~\frac{dy}{dx}=\frac{x+2}{(x-1)(x+5)}\left[\frac{1}{x+2}-\frac{1}{x-1}-\frac{1}{x+5}\right]~~\text{(ans.)}$

Alternatively,

$~~~~~y=\frac{x+2}{(x-1)(x+5)} \\ \text{or,}~~ y=\frac 12\left[\frac{2x+4}{(x-1)(x+5)}\right]\\ \text{or,}~~y=\frac 12\left[\frac{(x-1)+(x+5)}{(x-1)(x+5)}\right] \\ \text{or,}~~ y=\frac 12\left[\frac{1}{x+5}+\frac{1}{x-1}\right]\\ \therefore \frac{dy}{dx}\\=\frac12\left[-\frac{1}{(x+5)^2}-\frac{1}{(x-1)^2}\right]\\=-\frac 12\left[\frac{1}{(x+5)^2}+\frac{1}{(x-1)^2}\right]~~\text{(ans.)}$