Differentiate the Logarithmic functions:
π¦ = π₯^3 ππ^2 π₯ + π₯^2 πππ₯^3 + 1
yβ = 3x2Β ln2xΒ + x2Β + 2x. πππ₯3Β + 3x
π¦ = π₯3Β ππ2x+ π₯2Β πππ₯3Β + 1
y = x3Β ln2 x+ x2Β (3lnx) + 1
let x3Β ln2Β x= u and ln2 x= p, x3= q
then, uβ = pqβ + qpβ
qβ = 3x2
from p = ln2 x
let lnx = u
u'= 1/x
then p = u2
p' = 2u
therefore, p' = 2u. 1/x
p' = 2/x lnx
uβ = ln2Β x. 3x2Β + x3Β . 2/x lnx
uβ = 3x2Β ln2Β x+ 2x2 lnx
yβ = 3x2Β ln2 x+2x2Β lnx + 2x. πππ₯3Β + X2Β (3/x)
yβ = 3x2Β ln2Β x+ 2x2Β lnx+ 2x. πππ₯3Β + 3x
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