Find the differential coefficient of tan^-1x with respect to x.
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1
Answer:
Let f(x)=tan
−1
x and f(x+∂x)=tan
−1
(x+∂x)
Now,
dx
d
tan
−1
x=
∂x→0
∂x
tan
−1
(x+∂x)−tan
−1
x
Let t=tan
−1
x⇒tant=x
t+∂t=tan
−1
(x+∂x)
On putting values
if ∂x→0 and ∂t→0
dx
d
tan
−1
x=
∂x→0
tan(t+∂t)−
t+∂t−t
=
∂t→0
cos(t+∂t)
sin(t+∂t)
−
cost
∂t
=
∂t→0
(t+∂t)−(t+∂t)
∂t((t+∂t))
=
∂t→0
sin(t+∂t−t)
∂t[(t+∂t)]
∂t→0
(
sin(∂t)
∂t
)(t+∂t)=(1)
=
sec
2
t
1
=
1+tan
2
t
1
=
1+x
2
1
Anonymous:
Hope it helps
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