𝑓(𝑥) = 9 𝑙𝑜𝑔 (𝑥) + 12 ln (2𝑥 + 1)
Given "f(x)=9 log\u2061 x+12 ln\u2061(2x+1)"
Which may be written as
"f'(x)=d\/dx [9lnx\/ln10+12 ln\u2061(2x+1) ]"
"=9\/ln\u206110 (d(ln x))\/dx+12 d[ln\u2061(2x+1)]\/dx"
"= 9\/(xln 10)+12(2\/(2x+1))"
"=9\/(xln 10)+ 24\/(2x+1)"
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