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

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

Question Image
Grade:12

4 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
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
askIITians Faculty
milind
23 Points
6 years ago
milind
23 Points
6 years ago
milind
23 Points
6 years ago

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