Question
If y=(1+x)(1+x2)(1+x4)⋯(1+x2n) then dydx at x = 0 is
Answer: Option C
:
C
Since,y=(1+x)(1+x2)(1+x4)⋯(1+x2n),(1−x)y=(1−x2)(1+x2)(1+x4)⋯(1+x2n)=(1−x4)(1+x4)⋯(1+x2n+1)∴y=1−x2n+1(1−x)∵dydx=(1−x)(−2n+1.x2n)−(1−x2n+1)(−1)(1−x)2∴dydx∣∣x=0=(1)(0)−(1)(−1)(1)2=1
Was this answer helpful ?
:
C
Since,y=(1+x)(1+x2)(1+x4)⋯(1+x2n),(1−x)y=(1−x2)(1+x2)(1+x4)⋯(1+x2n)=(1−x4)(1+x4)⋯(1+x2n+1)∴y=1−x2n+1(1−x)∵dydx=(1−x)(−2n+1.x2n)−(1−x2n+1)(−1)(1−x)2∴dydx∣∣x=0=(1)(0)−(1)(−1)(1)2=1
Was this answer helpful ?
More Questions on This Topic :
Question 1. If y=√x+√y+√x+√y+....∞,then dydx is equal to....
Question 2. If √(x+y)+√(y−x)=a,then d2ydx2 equals....
Question 4. If sin (x+y)=loge(x+y),thendydx is equal to....
Question 5. If xcos y+ycos x=5. Then....
Question 10. The derivative of cos−1(x−1−xx−1+x) at x=−1 is....
Submit Solution