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JEE Main 2014
9 April
Online
Exam
JEE Main
Year
2014
Date
9 April
Mode
Online
Subject
Mathematics
Chapter
Differentiation
JEE Main 2014 · 9 April · Online · Mathematics · Differentiation
If y = e ^ nx , then d ^ 2 y d x ^ 2 . d ^ 2 x d y ^ 2 is equal to :
If
y
=
e
n
x
y = {e}^{nx}
y
=
e
n
x
, then
d
2
y
d
x
2
.
d
2
x
d
y
2
\dfrac{{d}^{2}y}{d{x}^{2}} . \dfrac{{d}^{2}x}{d{y}^{2}}
d
x
2
d
2
y
.
d
y
2
d
2
x
is equal to :
Options
A
n
e
−
n
x
n{e}^{- nx}
n
e
−
n
x
B
−
n
e
−
n
x
- n{e}^{- nx}
−
n
e
−
n
x
C
n
e
n
x
n{e}^{nx}
n
e
n
x
D
1
1
1
Correct answer
B.
−
n
e
−
n
x
- n{e}^{- nx}
−
n
e
−
n
x
Step-by-step solution
y = {e}^{nx} ⇒ \dfrac{dy}{dx} = n{e}^{nx}
{
∵
d
d
x
e
a
x
=
a
e
a
x
}
\left\{∵ \dfrac{d}{dx}{e}^{ax} = a{e}^{ax}\right\}
{
∵
d
x
d
e
a
x
=
a
e
a
x
}
⇒
d
2
y
d
x
2
=
n
2
e
n
x
⇒ \dfrac{{d}^{2}y}{d{x}^{2}} = {n}^{2}{e}^{nx}
⇒
d
x
2
d
2
y
=
n
2
e
n
x
d
y
d
x
=
n
e
n
x
⇒
d
x
d
y
=
1
n
e
n
x
=
1
n
e
−
n
x
\dfrac{dy}{dx} = n{e}^{nx} ⇒ \dfrac{dx}{dy} = \dfrac{1}{n{e}^{nx}} = \dfrac{1}{n}{e}^{- nx}
d
x
d
y
=
n
e
n
x
⇒
d
y
d
x
=
n
e
n
x
1
=
n
1
e
−
n
x
d
2
x
d
y
2
=
d
d
y
(
d
x
d
y
)
\dfrac{{d}^{2}x}{d{y}^{2}} = \dfrac{d}{dy}\left(\dfrac{dx}{dy}\right)
d
y
2
d
2
x
=
d
y
d
(
d
y
d
x
)
=
d
d
x
(
d
x
d
y
)
⋅
d
x
d
y
= \dfrac{d}{dx}\left(\dfrac{dx}{dy}\right) \cdot \dfrac{dx}{dy}
=
d
x
d
(
d
y
d
x
)
⋅
d
y
d
x
=
d
d
x
(
1
n
e
−
n
x
)
⋅
e
−
n
x
n
= \dfrac{d}{dx}\left(\dfrac{1}{n}{e}^{- nx}\right) \cdot \dfrac{{e}^{- nx}}{n}
=
d
x
d
(
n
1
e
−
n
x
)
⋅
n
e
−
n
x
=
1
n
(
−
n
)
e
−
n
x
⋅
e
−
n
x
n
= \dfrac{1}{n}\left(- n\right){e}^{- nx} \cdot \dfrac{{e}^{- nx}}{n}
=
n
1
(
−
n
)
e
−
n
x
⋅
n
e
−
n
x
=
−
1
n
e
−
2
n
x
= - \dfrac{1}{n}{e}^{- 2nx}
=
−
n
1
e
−
2
n
x
⇒
d
2
y
d
x
2
.
d
2
x
d
y
2
=
n
2
e
n
x
×
(
−
e
−
2
n
x
n
)
⇒ \dfrac{{d}^{2}y}{d{x}^{2}} . \dfrac{{d}^{2}x}{d{y}^{2}} = {n}^{2}{e}^{nx} \times \left(\dfrac{- {e}^{- 2nx}}{n}\right)
⇒
d
x
2
d
2
y
.
d
y
2
d
2
x
=
n
2
e
n
x
×
(
n
−
e
−
2
n
x
)
=
−
n
e
−
n
x
= - n{e}^{- nx}
=
−
n
e
−
n
x
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