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Differentiate 3x w.r.t. logx3.

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Question

Differentiate 3x w.r.t. logx3.

Sum
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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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Chapter 1: Differentiation - Exercise 1.4 [Page 49]

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