Differentiate the Logarithmic functions:
π¦ = π₯^3 ππ^2 π₯ + π₯^2 πππ₯^3 + 1
yβ = 3x2 ln2x + x2 + 2x. πππ₯3 + 3x
π¦ = π₯3 ππ2x+ π₯2 πππ₯3 + 1
y = x3 ln2 x+ x2 (3lnx) + 1
let x3 ln2 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 = ln2 x. 3x2 + x3 . 2/x lnx
uβ = 3x2 ln2 x+ 2x2 lnx
yβ = 3x2 ln2 x+2x2 lnx + 2x. πππ₯3 + X2 (3/x)
yβ = 3x2 ln2 x+ 2x2 lnx+ 2x. πππ₯3 + 3x
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