UPSC NDA Math Model Paper 2

Given: y=ex
To Find: (

d2x

dy2

)×(

d2y

dx2

)
y=ex
Differentiating with respect to x, we get
⇒

dy

dx

=ex
Again differentiating with respect to x, we get
⇒

d2y

dx2

=ex ......(1)
Again, y=ex
Taking log both sides, we get
⇒log‌y=log⇒ex
⇒log‌y=x‌log‌e (∵log‌mn=n‌log‌m)
⇒log‌y=x (∵log‌e=1)
⇒x=log‌y
Differentiating with respect to y, we get
⇒

dx

dy

=

1

y

Again differentiating with respect to y, we get
⇒

d2x

dy2

=

−1

y2

.......(2)
Multiplying equation (1) and (2), we get
(

d2x

dy2

)×(

d2y

dx2

)=

−1

y2

×ex
⇒(

d2x

dy2

)×(

d2y

dx2

)=

−1

y2

×y (∵y=ex)
∴(

d2x

dy2

)×(

d2y

dx2

)=

−1

y