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Question
Differentiate 3x w.r.t. log3x
Solution
Let `u=3^x`
`"du"/dx=3^xlog3`
Let `v=log_3x=logx/log3`
`"dv"/dx=1/log3d/dx(logx)`
`=1/log3 1/x=1/(xlog3)`
`"du"/"dv"=("du"/dx)/("dv"/dx)=(3^xlog3)/(1/(xlog3))=3^x.x(log3)`^2
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