i. ln[1+sin2x]
Differentiating with respect to x:
d[ln[1+sin2x]]/dx
=(1/(1+sin2x))∗[d(1+sin2x)/dx]
= (1/(1+sin2x))∗[d[1+(1−cos2x)/2]/dx]
=(1/(1+sin2x))∗[d(2+1−cos2x)/2]/dx]
=(1/2(1+sin2x))∗[2sin2x]
=sin2x/(1+sin2x) (Answer)
ii. xx
Let A=xx
Taking ln on both sides:
lnA=xlnx
Differentiating w.r.t x :
d[lnA]/dx=d[xlnx]/dx
=(1/A)∗d[A]/dx=x∗(1/x)+lnx
or,dA/dx=(1+lnx)∗A
=(1+lnx)∗xx (Answer)
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