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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

2 Answers

Arun Kumar IIT Delhi
askIITians Faculty 256 Points
6 years ago
Hello Student,

\\y=log^n x \\=>{dy \over dx}={1 \over log^{n-1}x*log^{n-2}x......logx} \\=>log^{n-1}x*log^{n-2}x......logx*{dy \over dx}=1 \\=>log^nx* log^{n-1}x*log^{n-2}x......logx*{dy \over dx}=log^nx
Thanks & Regards
Arun Kumar
Btech, IIT Delhi
Askiitians Faculty
milind
23 Points
6 years ago
Differential of log ^n x is 

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