wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If y=1+1x+1x2+1x3+..... with |x|>1 then dydx=

A
x2y2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
y2x2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
x2y2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
y2x2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D y2x2
y=1+1x+1x2+1x3+.....y=111x=xx1
dydx=(x1)x(x1)2=1(x1)2×x2x2
dydx=y2x2

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Logarithmic Differentiation
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon