1.i. ln[1+sin2x]
Differentiating with respect to x :
d(ln[1+sin2x])/dx
=[2cos(2x)]/[1+sin2x] (Answer)
ii. xx
Let A = xx
Taking natural logarithm on both sides:
lnA=xlnx
Differentiating with respect to x:
d[lnA]/dx=d[xlnx]/dx
or, 1/A∗d[A]/dx=x/x+lnx
or, dA/dx=(1+lnx)∗A
Putting A = xx
dA/dx=[1+lnx]∗XxAnswer
c. Solve ∫(2x3−4x−8/x4−x3+4x2−4x)dx
= 2x4/4−4x2/2−8x−3/(−3)−x4/4+4x3/3−4x2/2+c
where C is an integration constant
= x4/2−2x2+8(x−3/3)−x4/4+4x3/3−2x2+c
=[x4/2−x4/4]−[2x2+2x2]+8x−3/3+4x3/3+c
=[x4/4]−4x2+8/3x3+4x3/3+c[Answer]
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