CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

The derivative of tan1(1+x21x) with respect to tan1(2x1x212x2) at x=0 is

A
14
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
14
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 14
Let y=tan1(1+x21x) and z=tan1(2x1x212x2)
Putting x=tanθ in y, we get
y=tan1(secθ1tanθ)
y=tan1(1cosθsinθ)
y=tan1(tanθ2)=θ2
y=12tan1x
dydx=12(1+x2)

Putting x=sinθ in z, we get
z=tan1(2sinθcosθ12sin2θ)
z=tan1sin2θcos2θz=2θ=2sin1x
dzdx=21x2

dydz=1x24(1+x2)
at x=0, dydz=14

flag
Suggest Corrections
thumbs-up
0
BNAT
mid-banner-image