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

milind , 10 Years ago
Grade 12
anser 2 Answers
Arun Kumar

Last Activity: 10 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

Last Activity: 10 Years ago

Differential of log ^n x is 

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