2.1 DIFFERENTIATION
Find the derivative with respect toxoff(t)=2at,wherex=at^2.
We could of course substitute fortand then differentiatefas a function ofx, but in this
case it is quicker to use
df
dx
=
df
dt
dt
dx
=2a
1
2 at
=
1
t
,
where we have used the fact that
dt
dx
=
(
dx
dt
)− 1
.
2.1.4 Differentiation of quotients
Applying (2.6) for the derivative of a product to a functionf(x)=u(x)[1/v(x)],
we may obtain the derivative of the quotient of two factors. Thus
f′=
(u
v
)′
=u
(
1
v
)′
+u′
(
1
v
)
=u
(
−
v′
v^2
)
+
u′
v
,
where (2.12) has been used to evaluate (1/v)′. This can now be rearranged into
the more convenient and memorisable form
f′=
(u
v
)′
=
vu′−uv′
v^2
. (2.13)
This can be expressed in words asthe derivative of a quotient is equal to the bottom
times the derivative of the top minus the top times the derivative of the bottom, all
over the bottom squared.
Find the derivative with respect toxoff(x)=sinx/x.
Using (2.13) withu(x)=sinx,v(x)=xand henceu′(x)=cosx,v′(x) = 1, we find
f′(x)=
xcosx−sinx
x^2
=
cosx
x
−
sinx
x^2
.
2.1.5 Implicit differentiation
So far we have only differentiated functions written in the formy=f(x).
However, we may not always be presented with a relationship in this simple
form. As an example consider the relationx^3 − 3 xy+y^3 = 2. In this case it is
not possible to rearrange the equation to giveyas a function ofx. Nevertheless,
by differentiating term by term with respect tox(implicit differentiation), we can
find the derivative ofy.