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Question
Differentiate 3x w.r.t. logx3.
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Solution
Let u = 3x and v = logx3.
Then we want to find `"du"/"dv"`.
Differentiating u and v w.r.t. x, we get
`"du"/"dv" = "d"/"dx"(3^x)`
= 3x.log3
and
`"dv"/"dx" = "d"/"dx"(log_x3)`
= `"d"/"dx"((log3)/(logx))`
= `log3."d"/"dx"(logx)^-1`
= `(log3)(-1)(logx)^-2."d"/"dx"(logx)`
= `(-log3)/(logx)^2 xx (1)/x`
= `(-log3)/(x(logx)^2`
∴ `"du"/"dv" = (("du"/"dx"))/(("dv"/"dx")`
= `(3^x.log3)/([(-log3)/(x(logx)^2)]`
= – x(log x)2.3x.
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