Last Activity: 10 Years ago
y=(sinx)logx
logy=logx. Logsinx
(1/y)dy/dx=logx.(1/sinx)(cosx)+logsinx. (1/x)
dy/dx=sinxlogx[cotx.logx +(1/x)logsinx]
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Rinkoo Gupta
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Last Activity: 2 Years ago