If y=e^asin^-1x find value of y when x=a
shivanshusingh1992:
it's Yn and not y
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Answer:
To proceed we will need some standard Calculus results:
d
d
x
e
a
x
=
a
e
a
x
d
d
x
sin
−
1
x
=
1
√
1
−
x
2
Now we have:
y
=
e
m
sin
−
1
x
If we apply the chain rule then we get:
y
'
=
m
e
m
sin
−
1
x
⋅
1
√
1
−
x
2
=
m
e
m
sin
−
1
x
√
1
−
x
2
And differentiating again and applying the quotient rule, along with the chain rule, we get:
y
'
'
=
(
√
1
−
x
2
)
(
d
d
x
m
e
m
sin
−
1
x
)
−
(
m
e
m
sin
−
1
x
)
(
d
d
x
√
1
−
x
2
)
(
√
1
−
x
2
)
2
=
(
√
1
−
x
2
)
(
m
2
e
m
sin
−
1
x
√
1
−
x
2
)
−
(
m
e
m
sin
−
1
x
)
(
1
2
(
1
−
x
2
)
−
1
2
⋅
(
−
2
x
)
)
(
√
1
−
x
2
)
2
=
m
2
e
m
sin
−
1
x
+
m
x
e
m
sin
−
1
x
√
1
−
x
2
1
−
x
2
=
m
2
y
+
x
y
'
1
−
x
2
∴
(
1
−
x
2
)
y
'
'
=
m
2
y
+
x
y
'
∴
(
1
−
x
2
)
y
'
'
−
x
y
'
=
m
2
y
QED
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