Question

Differentiate 3^x w.r.t. log₃x

Answer 1

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