Understanding Engineering Mathematics

You should now be able to repeat this argument with the special case

ofa=2 to obtain

d

dx

( 2 x)= 2 xln 2

D.Ify=f(x)=

x− 1

x+ 2

, then we could use the quotient or product rule:

f′(x)=

1

x+ 2

d

dx

(x− 1 )+(x− 1 )

d

dx

(

1

x+ 2

)

=

1

x+ 2

−

x− 1

(x+ 2 )^2

So

f′( 1 )=

1

1 + 2

−

0

32

=

1

3

However it is much easier by implicit differentiation. Rewrite as:

(x+ 2 )y=x− 1

So, differentiating through with respect tox:

( 1 )y+(x+ 2 )y′= 1

Substitutingx=1andy=0gives

( 1 )( 0 )+ 3 y′( 1 )= 1

hence y′( 1 )=

1

3

as before.

8.2.6 Parametric differentiation

➤

229 245➤

Ifx,yare defined in terms of a parameter,t,(91
➤
)

x=x(t) y=y(t)

then the function of a function rule gives

dy

dx

=

dy

dt

dt

dx

Since

dt

dx

= 1

/

dx

dt

this can now be written as

dy

dx

=

dy/dt

dx/dt