Understanding Engineering Mathematics

Solution to review question 8.1.6

Withx= 3 t^2 ,y=cos(t+ 1 )we have

dy

dt

=−sin(t+ 1 )and

dx

dt

= 6 t,so:

dy

dx

=

dy/dt

dx/dt

=

−sin(t+ 1 )

6 t

8.2.7 Higher order derivatives

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229 245➤

In general,

dy

dx

will be a function ofx. We can therefore differentiate it again with respect

tox. We write this as

d

dx

(

dy

dx

)

=

d^2 y

dx^2

Note that

d^2 y

dx^2

doesnotmean

(

dy

dx

) 2

We can, of course, differentiate yet again and write

d

dx

(

d^2 y

dx^2

)

=

d^3 y

dx^3

and so on.

In general

dny

dxn

denotes thenth order derivative– differentiateyntimes successively.

Solution to review question 8.1.7

A. (i) A first differentiation gives

d

dx

( 2 x+ 1 )= 2

So, differentiating again

d^2

dx^2

( 2 x+ 1 )=

d

dx

( 2 )= 0

(ii) A bit quicker this time:

d^2

dx^2

(x^3 − 2 x+ 1 )=

d

dx

( 3 x^2 − 2 )= 6 x