3.7: Implicit Differentiation
P is on the curve: 22 + (−1)2 = 5.
Objective: Find dy
dx for curves given
y
by equations like x 2 + y 2 = 5 where y
is not given as a function y = f (x).
Technique:
I
1
Differentiate both sides of the
equation with respect to x.
Treat y as an unknown function
of x using dy
dx for its derivative.
dy
dx .
I
Then solve for
I
Slope of the curve at P(x0 , y0 ):
dy m=
dx (x0 ,y0 )
x
1
P
d 2
d 2
d
x +
y =
5
dx
dx
dx
dy
2x + 2y
=0
dx
dy
x
=−
dx
y
m = dy
= 2; so the tangent
dx 2
2
Example: Find dy
dx for x + y = 5.
(2,−1)
Then find an equation of the tangent
is
y
=
2(x
− 2) − 1 or y = 2x − 5.
line at the point P(2, −1).
We could have solved for y and
differentiated using the chain rule.
p
y = ± 5 − x2
dy
1
=± √
(−2x)
dx
2 5 − x2
x
x
=− .
= ∓√
y
5 − x2
Find the second derivative
d 2y
:
dx 2
y · 1 − x dy
d x d 2y
dx
=
−
=
−
dx 2
dx y
y2
y + x yx
y2 + x2
=−
=
−
y2
y3
−3/2
5
= − 3 = ∓5 5 − x 2
y
2
y
Find dy
dx for x = e and find an
equation of the tangent line at the
point P(1, 0).
d 2
d y
x =
e
dx
dx
dy
2x = e y
dx
2x
dy
2x
2
= y = 2 =
dx
e
x
x
m = dy
= 2; so the tangent is
dx (1,0)
y = 2(x − 1) + 0 or y = 2x − 2.
y
x 2 = ey
1
x
1
3
3
Find dy
dx for the curve x + y = 9xy .
Show that P(2, 4) lies on the curve
and find an equation of both the
tangent and normal lines at P.
(Example 5 of the text on p 167)
d 3
d 3
x +
y
dx
dx
dy
3x 2 + 3y 2
dx
dy
2
(3y − 9x)
dx
dy
dx
d
=
(9xy )
dx
dy
= 9y + 9x
dx
= 9y − 3x 2
9y − 3x 2
3y 2 − 9x
3y − x 2
= 2
y − 3x
=
Since 23 + 43 = 72 = 9 · 2 · 4,
P(2, 4) lies on the curve.
Slope of the curve mT at P(2, 4):
3 · 4 − 22
4
dy = 2
= .
mT =
dx (2,4) 4 − 3 · 2
5
So the tangent line has equation
y = 54 x + 12
5 .
The normal line is perpendicular to
the tangent; so mN = − m1T = − 54
and the normal line has equation
y = − 54 x + 13
2 .
√ Find the slopes at the 2 points on the Slopes of the curve at 0, ± 3 :
curve x 2 + xy + y 2 = 3, where it√ √
dy ± 3
1
crosses the y -axis. Points: 0, ± 3
m=
=− √ =− .
√
y
dx (0,± 3 )
2
±2 3
1
1
x
d 2
d
d 2
d
x +
(xy ) +
y =
3
dx
dx
dx
dx
dy
dy
2x + y + x
+ 2y
=0
dx
dx
dy
2x + y
=−
dx
x + 2y
If there’s time: Find dy
dx
y 3 = sin(2x + 5y )
d
d 3
(y ) =
(sin(2x + 5y ))
dx
dx
dy
dy = cos(2x + 5y ) 2 + 5
3y 2
dx
dx
dy
2 cos(2x + 5y )
= 2
dx
3y − 5 cos(2x + 5y )
Can we use implicit differentiation
to find the derivative of ln x? Yes!
dy
1
=
dx
x