Understanding Engineering Mathematics

so

2 x+ 2 y+ 2 x

dy

dx

+ 4 y

dy

dx

= 0

Hence(x+y)+(x+ 2 y)

dy

dx

= 0

and so

dy

dx

=−

x+y

x+ 2 y

B.This is a standard application of implicit differentiation. Ify=sin−^1 x

thenx=siny. But note that as usual with inverse trigonometric func-

tions we must restrict the range ofyto, say,−

π

2

≤y≤

π

2

to obtain

a single valued function (184

➤

).

Hence, on this range

dx

dy

=cosy

=

√

1 −sin^2 y

where we have taken the positive root since the cosine is positive on

−

π

2

≤y≤

π

2

=

√

1 −x^2

So, using

dy

dx

= 1

/

dx

dy

(see above) we get

dy

dx

=

1

√

1 −x^2

C.y=axis another case where we can use implicit differentiation to
advantage. If we take logs to baseewe have:

lny=lnax=xlna

Now differentiate through with respect tox:

d

dx

(lny)=

d

dx

(xlna)

or

1

y

dy

dx

=lna

by using the chain rule on the left-hand side.

So

dy

dx

=ylna=axlna