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# See attachment sir please explain this question …...I’ll be very great full

Jitender Singh IIT Delhi
6 years ago
Ans:
$y = log^{n}x$
$\frac{dy}{dx} = n.log^{n-1}x.\frac{1}{x}$
$L = x.logx.log^{2}x.log^{3}x.log^{4}x........log^{n}x.\frac{dy}{dx}$
$L = x.logx.log^{2}x.log^{3}x.log^{4}x........log^{n}x. n.log^{n-1}x.\frac{1}{x}$
$L = n.logx.log^{2}x.log^{3}x.log^{4}x........log^{n}x.log^{n-1}x$
$L = n.log^{(1+2+3+4+.........+n)}x.log^{n-1}x$
$L = n.log^{n(\frac{n+1}{2})}x.log^{n-1}x$
$L = n.log^{(\frac{n(n+1)}{2}+(n-1))}x$
Thanks & Regards
Jitender Singh
IIT Delhi
milind
23 Points
6 years ago
milind
23 Points
6 years ago
milind
23 Points
6 years ago