do now find the tangents to the curve at the points where the slope is 4. what is the smallest slope...
TRANSCRIPT
Do NowFind the tangents to the curve at the points wherethe slope is 4. What is the smallest slope of the curve? At whatvalue of x does the curve have this slope?
3y x x
Do NowFind the tangents to the curve at the points wherethe slope is 4. What is the smallest slope of the curve? At whatvalue of x does the curve have this slope?
3y x x
The derivative: 23 1y x x Find where the slope is 4:
23 1 4x 23 3 0x
3 1 1 0x x 1x
Slope 4, points (1,2) and (–1,–2).
Tangent lines: 4 2y x 4 2y x
For smallest slope, minimize
23 1y x x The smallest slope is 1,and occurs at x = 0.Graphical support???
As we learned last class, the derivative of the sum of twofunctions is the sum of their derivatives (and the same holdstrue for differences of functions). Is there a similar rule forthe product of two functions?
2f x xLet
2 2d dx x x x
dx dx The derivative:
11 1d dx x
dx dx However,
We need to derive a new rule for products…
f x u x v x Let
0
limh
u x h v x h u x v xduv
dx h
The derivative:
0
limh
u x h v x h u x h v x u x h v x u x v x
h
Subtract and add u(x + h)v(x) in the numerator:
0
limh
v x h v x u x h u xu x h v x
h h
0 0 0
lim lim limh h h
v x h v x u x h u xu x h v x
h h
f x u x v x Let
The derivative:
0 0 0
lim lim limh h h
v x h v x u x h u xu x h v x
h h
d du x v x v x u x
dx dx
Rule 5: The Product Rule
d dv duuv u v
dx dx dx
The product of two differentiable functions u and v isdifferentiable, and
To find the derivative of a product of twofunctions: “The first times the derivative of thesecond plus the second times the derivative ofthe first.”
u x
f xv x
How about when we have a quotient?...
0limh
u x h u x
v x h v xd u
dx v h
The derivative:
Subtract and add v(x)u(x) in the numerator:
0
limh
v x u x h u x v x h
hv x h v x
0
limh
v x u x h v x u x v x u x u x v x h
hv x h v x
u x
f xv x
How about when we have a quotient?...
The derivative:
0
limh
v x u x h v x u x v x u x u x v x h
hv x h v x
0limh
u x h u x v x h v xv x u x
h hv x h v x
2d d
v x u x u x v xdx dx
v x
Rule 6: The Quotient Rule
2
du dvv ud u dx dx
dx v v
At a point where , the quotient of twodifferentiable functions is differentiable, and
To find the derivative of a quotient of twofunctions: “The bottom times the derivative ofthe top minus the top times the derivative ofthe bottom, all divided by the bottom squared.”
0v y u v
Practice Problems
2 1u x
Find if f x 2 31 3f x x x
3 3v x Let’s use the product rule with
and
2 31 3d
f x x xdx
2 2 31 3 3 2x x x x 4 2 43 3 2 6x x x x 4 25 3 6x x x Any other method for
finding this answer?
Practice Problems
2 1u x
Differentiate 2
2
1
1
xf x
x
2 1v x Use the quotient rule with and
2 2
22
1 2 1 2
1
x x x xf x
x
3 3
22
2 2 2 2
1
x x x x
x
22
4
1
x
x
Graphical support: 1y f x 2 NDERy f x
Practice Problems
Let be the product of the functions u and v.y uvFind if 2y
2 3u 2 4u 2 1v 2 2v
From the Product Rule: y uv uv vu At our particular point:
3 2 1 4
2 2 2 2 2y u v v u
2
Practice Problems
Suppose u and v are functions of x that are differentiable at x = 2.Also suppose that
2 3u 2 4u 2 1v 2 2v Find the values of the following derivatives at x = 2.
(a)d u
dx v
2
2 2 2 2
2
v u u v
v
2
1 4 3 2
1
10
Practice Problems
Suppose u and v are functions of x that are differentiable at x = 2.Also suppose that
2 3u 2 4u 2 1v 2 2v Find the values of the following derivatives at x = 2.
(b)d v
dx u
2
2 2 2 2
2
u v v u
u
2
3 2 1 4
3
10
9