Understanding Engineering Mathematics

(やまだぃちぅ) #1
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



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
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