What is the derivative of x32log2x+1?

1 Answer
Mar 13, 2018

x2ln2[xx+1+3lnx+1]

Explanation:

Note:(1)logab=logeblogea
(2)ddx(uv)=udvdx+vdudx
y=x32log2x+1=x32[logex+1loge2]
y=1loge2x32logex+1
y=1ln2x32lnx+1
dydx=1ln2[x32ddx(lnx+1)+lnx+1ddx(x32)]
=1ln2[x321x+112x+1+lnx+132x12]
=1ln2[xx2(x+1)+32xlnx+1]
=x2ln2[xx+1+3lnx+1]