This week
Tangent lines to a circle
Find the slope of the tangent line to the circle
(x − 2)2 + (y + 3)2 = (13)2
Homework #5 due Thursday 11:30
Read sections 3.5 and 3.6
at the point (7, 9) using the following techniques.
Worksheet #5 on Tuesday
tangent line is perpendicular to the radial line.
solve for y and differentiate
implicit differentiation
Professor Christopher Hoffman
Math 124
Using the radial line
Professor Christopher Hoffman
Math 124
Solving for y
Solving for y we get that
(y + 3)2 = 169 − (x − 2)2 .
The center of the circle is (2, −3). The slope of the line
between the center and (7, 9) is
mradial
9 − (−3)
12
=
=
.
7−2
5
The slope of the tangent line is given by
mtangent =
−1
−1
−5
=
=
.
mradial
12/5
12
Professor Christopher Hoffman
Math 124
y = −3 ±
q
169 − (x − 2)2 .
As (7, 9) is on the top half of the circle we have that
q
y = −3 + 169 − (x − 2)2 .
Differentiating (and using the chain rule twice)
1
y 0 = 0 + (169 − (x − 2)2 )−1/2 (−2(x − 2))(1)
2
Evaluated when x = 7 we get
1
−5
−5
y 0 = (169 − (7 − 2)2 )−1/2 (−2(7 − 2)) = √
=
.
2
12
144
Professor Christopher Hoffman
Math 124
Implicit Differentiation
Implicit Differentiation
Now we treat y as a function of x and differentiate both sides
(remembering to use the chain rule!). To emphasize that y is a
function of x we write y (x).
(y (x) + 3)2 + (x − 2)2 = 169.
d d
(y (x) + 3)2 + (x − 2)2 =
(169) .
dx
dx
2(y (x) + 3)y 0 (x) + 2(x − 2) = 0.
Evaluating
y 0 (x) =
−2(x − 2)
.
2(y (x) + 3)
at the point (7, 9) we get
y 0 (7) =
−2(7 − 2)
−5
=
.
2(9 + 3)
12
Now we solve for y 0 in terms of x and y .
y 0 (x) =
−2(x − 2)
.
2(y (x) + 3)
Professor Christopher Hoffman
Math 124
Professor Christopher Hoffman
Implicit Differentiation
Use implicit differentiation to find
Math 124
dy
dx
if
4y 2 + x 2 = 4
√
Find the tangent line at the point at (1, − 3/2).
Treat y as a function of x.
Differentiate both sides. (Use the chain rule!).
Solve for y 0 in terms of x and y .
Plug in values of x and y to find y 0
Professor Christopher Hoffman
Math 124
4y 2 + x 2
d 2
4y + x 2
dx
dy
8y
+ 2x
dx
dy
dx
dy
dx
Professor Christopher Hoffman
=
=
=
=
=
Math 124
4
d
(4)
dx
0
−2x
8y
−x
4y
Looking
√ at the graph we see the ellipse and its tangent line at
(1, − 3/2).
√
At the point (1, − 3/2) we get
dy
−2x
−2(1)
1
√
=
=
= √ .
dx
8y
8(− 3/2)
2 3
So the equation of the tangent line is
√
(y +
1
3/2) = √ (x − 1).
2 3
Professor Christopher Hoffman
Math 124
Suppose that
x 3 + y 3 = 6xy .
Find
dy
dx .
Find the equation of the tangent line at (3, 3).
At which points on the curve do we have a vertical tangent
line?
Professor Christopher Hoffman
Math 124
Professor Christopher Hoffman
x3 + y3
d 3
x + y3
dx
dy
3x 2 + 3y 2
dx
dy 2
3y − 6x
dx
dy
dx
dy
dx
Professor Christopher Hoffman
Math 124
=
=
=
=
=
=
6xy
d
(6xy )
dx
dy
6 x
+ y (1)
dx
6y − 3x 2
6y − 3x 2
3y 2 − 6x
2y − x 2
.
y 2 − 2x
Math 124
Find the equation of the tangent line at (3, 3).
The slope of the tangent line is
2y − x 2
.
y 2 − 2x
Plugging in x = 3 and y = 3 we get
m=
−3
2(3) − (3)2
=
= −1.
3
(3)2 − 2(3)
Using the point slope formula the tangent line is
y − 3 = −1(x − 3).
At which points on the curve do we have a vertical tangent
line?
The slope of the tangent line is
2y − x 2
y 2 − 2x
If the tangent line is vertical then the denominator must be 0.
Thus y 2 − 2x = 0 or x = y 2 /2.
When we plug this into the equation
x 3 + y 3 = 6xy
we get
(y 2 /2)3 + y 3 = 6(y 2 /2)y
y 6 + 8y 3 = 24y 3
y 6 − 16y 3 = 0
Professor Christopher Hoffman
Math 124
Professor Christopher Hoffman
Math 124
y 3 (y 3 − 16) = 0
there are two tangent lines at (0, 0) and one of them is vertical.
3
3
y (y − 16) = 0.
Thus either y 3 = 0 or y 3 = 16.
If y 3 = 0 then y = 0 and x = 0.
If y 3 = 16 then y = 24/3 , and x = 12 y 2 = 25/3 .
Thus the two candidates for a vertical tangent line occur at
(0, 0) and (25/3 , 24/3 ).
At (25/3 , 24/3 )
dy
2(24/3 ) − (25/3 )2
27/3 − 210/3
= 4/3 2
=
dx
0
(2 ) − 2(25/3 )
At (0, 0) we get
dy
dx
=
0
0
so we need to check something else.
Professor Christopher Hoffman
Math 124
Professor Christopher Hoffman
Math 124
there is a vertical tangent line x = 25/3 .
Professor Christopher Hoffman
and the line y − 3 = −(x − 3) is tangent to the graph at (−3, 3)
Math 124
Professor Christopher Hoffman
Math 124
Find all points on the curve
dy
−y − 2xy 2
=
dx
2x 2 y + x
x 2 y 2 + xy = 2
where the slope of the tangent line is −1.
First we will find dy
dx . Then we will find all solutions to the
simultaneous equations
x 2 y 2 + xy = 2
and
dy
= −1.
dx
Now we set
dy
dx
= −1.
−1 =
dy
−y − 2xy 2
=
dx
2x 2 y + x
d 2 2
d
x y + xy =
(2)
dx
dx
2x 2 y + x = y + 2xy 2
dy
dy
x 2 (2y ) + y 2 (x 2 )0 + x
+ y (x)0 = 0
dx
dx
2x 2 y + x − y − 2xy 2 = 0
x 2 (2y
dy
dy
) + 2xy 2 + x
+y =0
dx
dx
dy
(2x 2 y + x) = −y − 2xy 2
dx
Professor Christopher Hoffman
Math 124
(x − y )(1 + 2yx) = 0
x =y
or
Professor Christopher Hoffman
2yx = −1
Math 124
Case 2: xy = −1/2
Case 1: x = y
x 2 y 2 + xy = 2
x 2 x 2 + xx = 2
If xy = −1/2 then
x4 + x2 − 2 = 0
(x 2 − 1)(x 2 + 2) = 0
x2 − 1 = 0
x 2 y 2 + xy = (−1/2)2 + (−1/2) = −1/4 6= 2.
This case yields no points.
Thus there are two points on the curve where the slope of the
tangent line is −1, (−1, −1) and (1, 1).
x = ±1
Thus we have two points
(−1, −1)
Professor Christopher Hoffman
and
(1, 1).
Math 124
Looking at the graph we that at (−1, −1) and (1, 1) the slope of
the tangent line is −1.
Professor Christopher Hoffman
Math 124
Implicit Differentiation
Treat y as a function of x.
Differentiate both sides. (Use the chain rule!).
Solve for y 0 in terms of x and y .
Plug in values of x and y to find y 0
Professor Christopher Hoffman
Math 124
Professor Christopher Hoffman
Math 124
Find the derivative of
y 5 + 3x 2 y 2 + 5x 4 = 93
Find the equation of the tangent line at the point (2, 1).
d 5
d
y + 3x 2 y 2 + 5x 4 =
(93)
dx
dx
5y 4
dy
dy
+ 3 x 2 2y
+ y 2 2x + 20x 3 = 0
dx
dx
(5y 4 + 6x 2 y )
dy
= −6y 2 x − 20x 3
dx
Find the equation of the tangent line at the point (2, 1).
dy
−6y 2 x − 20x 3
−6(1)2 (2) − 20(2)3
−172
=
=
.
=
dx
29
5y 4 + 6x 2 y
5(1)4 + 6(2)2
The equation of the tangent line at the point (1, 2) is
y −1=
−172
(x − 2).
29
dy
−6y 2 x − 20x 3
=
dx
5y 4 + 6x 2 y
Professor Christopher Hoffman
Math 124
Professor Christopher Hoffman
Math 124
At how many points does the curve
2y 3 + y 2 − y 5 = x 4 − 2x 3 + x 2
have a horizontal tangent line. Find the x coordinates of those
points.
Professor Christopher Hoffman
Math 124
Professor Christopher Hoffman
Math 124
d
d
2y 3 + y 2 − y 5 =
x 4 − 2x 3 + x 2
dx
dx
dy
dy
dy
+ 2y
− 5y 4
= 4x 3 − 6x 2 + 2x
dx
dx
dx
dy
6y 2 + 2y − 5y 4 = 4x 3 − 6x 2 + 2x
dx
6y 2
dy
4x 3 − 6x 2 + 2x
=
dx
6y 2 + 2y − 5y 4
dy
2x(2x 2 − 3x + 1)
=
dx
6y 2 + 2y − 5y 4
dy
2x(2x − 1)(x − 1)
=
dx
6y 2 + 2y − 5y 4
The horizontal tangents and occur when x = 0, x = 1/2 and
x = 1.
Professor Christopher Hoffman
Math 124
Professor Christopher Hoffman
Use implicit differentiation to find the derivative of y = sin−1 (x).
Taking the sin of both sides we get
sin(y ) = x.
Then
Math 124
dy
1
=
dx
cos(y )
Now we write cos(y ) in terms of x.
cos2 (y ) + sin2 (y ) = 1
d
d
(sin(y )) =
(x).
dx
dx
cos(y )
dy
= 1.
dx
dy
1
1
=
=
.
dx
cos(y )
cos(sin−1 (x))
we have
cos (y ) = 1 − sin2 (y ) = 1 − x 2
2
and
cos(y ) =
Then
p
1 − x 2.
dy
1
1
=
=√
.
dx
cos(y )
1 − x2
Let’s find a nicer formula.
Professor Christopher Hoffman
Math 124
Professor Christopher Hoffman
Math 124
Inverse Trig Functions
Use implicit differentiation to find the derivative of y = tan−1 (x).
Taking the tan of both sides we get
tan(y ) = x.
Then
d
d
(tan(y )) =
(x).
dx
dx
sec2 (y )
dy
= 1.
dx
cos2 (y )
1
1
cos2 (y )
dy
=
=
=
.
=
1
dx
1 + x2
1 + tan2 (y )
cos2 (y ) + sin2 (y )
Professor Christopher Hoffman
Math 124
d −1 1
.
sin (x) = √
dx
1 − x2
−1
d cos−1 (x) = √
.
dx
1 − x2
d −1 1
tan (x) =
.
dx
1 + x2
d −1
csc−1 (x) = √
.
dx
x x2 − 1
d 1
−1
sec (x) = √
.
dx
x x2 − 1
d −1 −1
cot (x) =
.
dx
1 + x2
Professor Christopher Hoffman
Math 124
Find the derivative of
f (x) = x tan−1 (4x).
√
Find f 0 (x) when f (x) = x sin−1 ( x).
√
√
f 0 (x) = x(sin−1 ( x))0 + (x)0 sin−1 ( x)
f 0 (x) = x(tan−1 (4x))0 + (x)0 tan−1 (4x)
√
1
1
= xp
( x −1/2 ) + sin−1 ( x).
√
1 − ( x)2 2
1
=x
(4) + tan−1 (4x).
1 + (4x)2
Professor Christopher Hoffman
Math 124
Professor Christopher Hoffman
Math 124
Use implicit differentiation to find the derivative of y = ln(x).
First we exponentiate both sides to get
We also have that if y = loga (x) then
ay = aloga (x) = x.
ey = eln(x) = x.
Taking derivatives we get
d
d
(ey ) =
(x) .
dx
dx
Taking derivatives
d y
d
(a ) =
(x).
dx
dx
dy
e
= 1.
dx
y
ln(a)ay
dy
= 1/ey .
dx
dy
= 1/x.
dx
dy
= 1/ ln(a)x.
dx
d
(ln(x)) = 1/x.
dx
Professor Christopher Hoffman
Math 124
dy
= 1.
dx
dy
= 1/ ln(a)ay .
dx
Professor Christopher Hoffman
Math 124
Find the derivative of
g(x) = ln(8x 6 + 3x 2 )
Find the derivatives of
By the chain rule
m(x) = ln
1
g (x) =
(8x 6 + 3x 2 )0
8x 6 + 3x 2
√
cos(x)(4x − 3)7 x + 3
0
1
=
(48x 5 + 6x).
8x 6 + 3x 2
For any function f (x) the chain rule tells us
It is best to use our rules of logarithms before taking derivatives.
1
m(x) = ln cos(x) + 7 ln(4x − 3) + ln x + 3)
2
m0 (x) =
1
1
1
1
(− sin(x)) + 7
(4) +
.
cos(x)
4x − 3
2 (x + 3)
d
1 0
ln(f (x)) =
f (x).
dx
f (x)
Professor Christopher Hoffman
Math 124
Professor Christopher Hoffman
Math 124
Logarithmic Differentiation
The last problem suggests a useful trick. Suppose
√
g(x) = cos(x)(4x − 3)7 x + 3.
so m(x) = ln(g(x)).
Then by the chain rule
m0 (x) =
1 0
g (x)
g(x)
and
g 0 (x) = g(x)m0 (x).
the function has a product or quotient of a large number of
terms or
In this case
0
0
g (x) = g(x)m (x) = cos(x)(4x − 3)
7
√
x +3
1
1
1
1
(− sin(x)) + 7
(4) +
cos(x)
4x + 3
2 (x + 3)
Professor Christopher Hoffman
Sometime it is necessary (or easier) to take natural logs before
taking the derivative. Then we proceed with implicit
differentiation.
We use this when
there is an x in both the base and exponent.
.
Professor Christopher Hoffman
Math 124
Math 124
Power Rule
Differentiate
√
x 3/4 x 2 + 1
y=
.
(3x + 2)5 (2x + 1)
Taking natural log of both sides we get
ln(y ) =
3
1
ln(x) + ln(x 2 + 1) − 5 ln(3x + 2) − ln(2x + 1).
4
2
d
d
(ln(y )) =
dx
dx
3
1
ln(x)+ ln(x 2 +1)−5 ln(3x +2)−ln(2x +1) .
4
2
1 0 3
1 2x
1
1
y = (1/x) +
−5
(3) −
(2).
y
4
2 x2 + 1
3x + 2
2x + 1
√
3
1 2x
1
1
x 3/4 x 2 + 1
0
y =
(1/x)+ 2
−5
(3)−
(2)
.
4
2x +1
3x + 2
2x + 1
(3x + 2)5 (2x + 1)
Professor Christopher Hoffman
Math 124
Prove the power rule by logarithmic differentiation.
If f (x) = x n then
ln(f (x)) = ln(x n ) = n ln(x).
Taking derivatives we get
d
d
(ln(f (x))) =
(n ln(x)) .
dx
dx
1 0
n
f (x) = .
f (x)
x
f 0 (x) = f (x)
n
n
= x n = nx n−1 .
x
x
Professor Christopher Hoffman
Math 124
Derivatives of functions with exponents
Use logarithmic differentiation to find the derivative of y = x x .
If a and b are constants then
d ln(y ) = x ln(x).
d
d
(ln(y )) =
(x ln(x)) .
dx
dx
dx
d f (x) a
= ln(a)af (x) f 0 (x).
dx
1 dy
1
= x + ln(x)(1).
y dx
x
d f (x)b = bf (x)b−1 f 0 (x).
dx
dy
= y (1 + ln(x)).
dx
d f (x)g(x) use logarithmic differentiation
dx
dy
= x x (1 + ln(x)).
dx
Professor Christopher Hoffman
Math 124
Professor Christopher Hoffman
If
√
2
If y = (2 x)3x +2x find y 0 .
y=
√
ln(y ) = ln((2 x)
3x 2 +2x
).
√
ln(y ) = (3x + 2x) ln(2 x).
2
√
1 dy
1 1
= (3x 2 + 2x) √ 2 x −1/2 + (6x + 2) ln(2 x).
y dx
2 x 2
√
dy
1
1
= y (3x 2 + 2x) √ 2 x −1/2 + (6x + 2) ln(2 x) .
dx
2 x 2
√
√
dy
1 1
2
= (2 x)3x +2x (3x 2 + 2x) √ 2 x −1/2 + (6x + 2) ln(2 x)) .
dx
2 x 2
Professor Christopher Hoffman
Math 124
ab = 0.
Math 124
√
(2x − 5) x 2 + 3(x 4 + 1)2.3
(x − 2)(x 5 − 2)(3 − 7x)x 4
find y 0 .
ln(y ) = ln
ln(y ) = ln(2x − 5) +
!
√
(2x − 5) x 2 + 3(x 4 + 1)2.3
.
(x − 2)(x 5 − 2)(3 − 7x)x 4
1
ln(x 2 + 3) + 2.3 ln(x 4 + 1) − ln(x − 2)
2
− ln(x 5 − 2) − ln(3 − 7x) − 4 ln(x).
d
d
(ln(y )) =
dx
dx
1
ln(2x−5)+ ln(x 2 +3)+2.3 ln(x 4 +1)−ln(x−2)
2
− ln(x 5 − 2) − ln(3 − 7x) − 4 ln(x) .
Professor Christopher Hoffman
Math 124
y (x) = (5 + 2x 2 )
p
x 4 + 3(x 4 + 1)x .
Find y 0 (x).
1 dy
1
1 1
1
1
=
2+
2x + 2.3 4
4x 3 −
(1)
y dx
2x − 5
2 x2 + 3
x −2
x +1
1
1
1
5x 4 −
(−7) − 4 .
x5 − 2
3 − 7x
x
!
√
4
2.3
2
1
dy
(2x − 5) x + 3(x + 1)
1 1
=
2+
2x
dx
2x − 5
2 x2 + 3
(x − 2)(x 5 − 2)(3 − 7x)x 4
1
1
1
1
1
(1) − 5
5x 4 −
(−7) − 4
+2.3 4
4x 3 −
.
x −2
x −2
3 − 7x
x
x +1
−
Professor Christopher Hoffman
Math 124
p
ln(y (x)) = ln (5 + 2x 2 ) x 4 + 3(x 4 + 1)x .
1
ln(x 4 + 3) + x ln(x 4 + 1).
2
1
2
ln(5 + 2x ) + ln(x 4 + 3) + x ln(x 4 + 1) .
2
ln(y (x)) = ln(5 + 2x 2 ) +
d
d
(ln(y (x))) =
dx
dx
1 dy
1
1 1
1
=
(4x)+ 4
(4x 3 )+x 4
(4x 3 )+(1) ln(x 4 +1).
y (x) dx
2x +3
5 + 2x 2
x +1
p
dy
= (5 + 2x 2 ) x 4 + 3(x 4 + 1)x
dx
1
1 1
1
(4x) +
+x 4
(4x 3 ) + (1) ln(x 4 + 1) .
2 x4 + 3
5 + 2x 2
x +1
Professor Christopher Hoffman
Math 124
Related Rates
Let A(t) be the area of the spill.
Let r (t) be the radius of the spill.
An oil platform explodes and spills oil into the Gulf of Mexico.
The oil spill grows in a circular fashion. If the radius of the spill
is growing at a rate of 25 kilometers per day when the radius is
220 kilometers, how fast is the area of the spill growing?
Professor Christopher Hoffman
Math 124
A(t) = π(r (t))2 .
0
A (t) = 2πr (t)r 0 (t).
Professor Christopher Hoffman
Math 124
Strategy
Let t0 be the time when r (t0 ) = 220. We are asked to find A0 (t0 ).
We have that
r (t0 ) = 220
and
r 0 (t0 ) = 25.
1
Draw a picture.
2
Choose notation.
3
Write an equation.
4
Differentiate using the chain rule.
5
Substitute into the equation and solve.
Plugging into
0
0
A (t) = 2πr (t)r (t)
we get
0
A (t) = 2π(220)(25) = 11000π.
Professor Christopher Hoffman
Math 124
Professor Christopher Hoffman
Math 124