sin y + cos x = x + 1, which gives rise to a
(x, y) = (0, 0) is a solution to that equation, and
there is a line tangent to the curve at (0, 0).
We examined the equation
in the plane. Note that
curve is nice enough,
curve
if the
Even though we are not able to write down a formula we know how to diery in terms of x near (0, 0) (in fact, y = sin−1 (1 + x − cos x) works for
entiate for
y
between
−π/2
π/2),
and
we can use the method of implicit dierentiation
to have the chain rule tell us the slope of the tangent line at
(0, 0).
sin y + cos x = x + 1
⇒
⇒
d
d
(sin y + cos x) =
(x + 1)
dx
dx
d
d
d
d
sin y +
cos x =
x+
1
dx
dx
dx
dx
⇒
⇒ (cos y)
(outer function is
d
sin y − sin x = 1 + 0
dx
d
y − sin x = 1
dx
f (x) = sin x,
⇒
by the chain rule
inner function is
g (x) = y
near
(0, 0) )
1 + sin x
d
y=
dx
cos y
Therefore,
dy 1 + sin 0
=
=1
dx (x,y)=(0,0)
cos 0
Since the slope of the tangent line through
line is
y − 0 = 1 (x − 0),
or
y = x.
(0, 0)
is
1,
the equation of the
The curve together with this tangent line
looks like the following:
1
0
