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Differentiate the functions given in w.r.t. $x :$ $ \sqrt {\large\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} $

$\begin{array}{1 1} \large\frac{1}{2}\sqrt {\large\frac{(x-1)(x-2)}{(x-8)(x-4)(x-7)}}\begin{bmatrix}\large\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\end{bmatrix} \\\large\frac{1}{2}\sqrt {\large\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\begin{bmatrix}\large\frac{1}{x+1}+\frac{1}{x-2}-\frac{1}{x+3}-\frac{1}{x-4}-\frac{1}{x-5}\end{bmatrix} \\ \large\frac{1}{2}\sqrt {\large\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\begin{bmatrix}\large\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\end{bmatrix} \\ \large\frac{1}{4}\sqrt {\large\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\begin{bmatrix}\large\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\end{bmatrix} \end{array} $

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1 Answer

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Toolbox:
  • $\log m^n=n\log m$
  • $\log mn=\log m+\log n$
  • $\log \large\frac{m}{n}$$=\log m-\log n$
Step 1:
Let $y= \sqrt {\large\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} $
Taking $\log$ on both sides we get
$\log y=\log \sqrt {\large\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} $
$\log y=\log \begin{bmatrix}\large\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)} \end{bmatrix}^{\large\frac{1}{2}}$
$\;\;\;=\large\frac{1}{2}$$\big(\log[(x-1)(x-2)]-\log[(x-3)(x-4)(x-5)]\big)$
$\;\;\;=\large\frac{1}{2}\big($$[\log(x-1)+\log(x-2)]-[\log(x-3)+\log(x-4)+\log(x-5)]\big)$
We know that $\log mn=\log m+\log n$
$\Rightarrow \large\frac{1}{2}$$[\log(x-1)+\log(x-2)-\log(x-3)-\log(x-4)-\log(x-5)]$
Step 2:
Differentiating on both sides we get
$\large\frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\begin{bmatrix}\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\end{bmatrix}$
$\large\frac{dy}{dx}=y.\frac{1}{2}\begin{bmatrix}\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\end{bmatrix}$
$\quad\;=\large\frac{1}{2}$$\sqrt {\large\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\begin{bmatrix}\large\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\end{bmatrix}$
answered May 8, 2013 by sreemathi.v
 

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