Eeman Faisal
Assignment Derivative Rules due 10/24/2022 at 05:00pm PKT
Math 101 Fall 22 23
Problem 1. (1 point)
Problem
Problem 2. (1 point)
Problem
Proof of the product rule: Suppose that each of g and h is a function with domain all numbers so that each of g0 and h0 have domain all numbers. Suppose also that f is the function so that
f (x) = g(x)h(x) for all numbers x (that is, f = gh). Then
Application of the product rule.
f (p) f (a)
= g(
p a
Suppose that g is differentiable at x and c is real number. Let
f = gh, where h is the function h(x) = c for all x. Then the product rule gives
) h(p)p h(a)
+ h(
a
The rule for constant multiples is a special case of the product
rule.
) g(p)p g(a)
a
d
As p ! a, the first quotient on the right goes to dx
x=a
d
As p ! a, the second quotient on the right goes to dx
x=a
(x)
f 0 (x) = g0 (x)h(x) + g(x)h0 (x)
(x)
But, h0 (x) =
We conclude that
f 0 (a) = g(a)h0 (a) + h(a)g0 (a)
so
Answer(s) submitted:
• a
• a
• h
• g
f 0 (x) = g0 (x)h(x)+
and using the definition of h(x) gives
f 0 (x) =
(correct)
g0 (x)
Answer(s) submitted:
• 0
• 0
• c
(correct)
1
Problem 3. (1 point)
Problem
Problem 4. (1 point)
Problem
Product rule and rule for positive integer powers of x:
For f differentiable at x, use mathematical induction to show that
d n
Consider f (x) = xn . We would like to prove that dx
x = nxn 1 ,
for n a positive integer.
d n
n 1 f0
dx f (x) = n f
for n = 1, 2, 3, . . . .
d n
Let Pn be the proposition that dx
x = nxn 1 , n a positive integer.
Enter 99 when done.
d 1
P1 is the proposition that dx
x = 1 · x1 1 = x0 = 1.
Answer(s) submitted:
• 99
We already checked this, so P1 is true.
d k
Pk is the proposition that dx
x =
.
k 1
(correct)
.
Suppose Pk
d k+1
Can we prove Pk+1 , the proposition that dx
x
=
Problem 5. (1 point)
Problem
xk ?
d k+1
d
d
d
LHS = dx
x
= dx
x · xk = x dx
(xk ) + xk dx
(x) by the product
rule.
For f differentiable at x,
Then by applying P1 and Pk we get
d n
n 1 f 0 (x)
dx f (x) = n f
LHS = x(kxk 1 ) + xk = kxk + xk = (k + 1)xk = RHS
for n = 1, 2, 3, . . . .
So, Pk =) Pk+1
d
dx
⇥ a
⇤
cos (x) + x2 + 7 =
cosa 1 (x) sin(x)+
x
Answer(s) submitted:
• -a
• 2
Because P1 and Pk =) Pk+1 we conclude Pn by mathematical induction.
(correct)
Answer(s) submitted:
• k
• x
• k+1
Problem 6. (1 point)
Problem
(correct)
Find the derivative of
1
f (x) = x 3 6x3 + 6x2 + 6
The only value of x for which the derivative is undefined is x =
Answer(s) submitted:
• 0
(correct)
2
Problem 7. (1 point)
Problem
Problem 8. (1 point)
Problem
We proved
We saw that for f differentiable at x and f (x) 6= 0 that
d
dx
⇣ ⌘
1
f
(x) =
f 0 (x)
f 2 (x)
d
dx
from the limit definition of derivative. We will also need the product rule for the following problem:
d
d
dx tan(x) = dx
d
dx tan(x) =
h
i
1
sin(x) · cos(x)
1
(x) · cos(x)
+ sin(x)·
✓ ◆
1
(x) =
f
f 0 (x)
[ f (x)]2
We also have the product rule.
These results are sufficient to prove the quotient rule:
(x)/ cos2 (x)
For f and g differentiable at x and g(x) 6= 0
d
2
dx tan(x) = sec (x)
d
dx
Answer(s) submitted:
• cos
• sin
✓ ◆
f
g(x) f 0 (x) f (x)g0 (x)
(x) =
g
g2 (x)
Proof using the product and reciprocal rules.
(correct)
d
dx
d
dx
d
dx
⇣ ⌘
f
g
⇣ ⌘
f
g
⇣ ⌘
f
g
d
(x) = dx
[ f (x) · 1/
(x) =
f (x)·
(x) = ( f 0 (x)g(x)
Answer(s) submitted:
• g
• g
• g
• f
(correct)
3
(x)]
(x)/[g(x)]2 + f 0 (x)/
(x)g0 (x)) /[g(x)]2
(x)
Problem 9. (1 point)
Problem
Problem 10. (1 point)
Problem
Let’s find the derivative of cotangent.
Recall the quotient rule:
Recall the quotient rule:
For f , g differentiable at x and g(x) 6= 0 we have
For f , g differentiable at x and g(x) 6= 0 we have
d
dx
d
dx
✓ ◆
f
g(x) f 0 (x) f (x)g0 (x)
(x) =
g
g2 (x)
Yes/No
cos(x)·
Answer(s) submitted:
• No
(x)) / sin (x)
/ sin (x) =
0
Are there any values of x for which the derivative is not defined?
2
2
0
f (x)g (x)
(x) = g(x) f (x)
g2 (x)
d x2 + x + 1
(1 + sin2 (x))(2x + 1) (x2 + x + 1)(2 sin(x) cos(x))
=
2
dx 1 + sin (x)
(1 + sin2 (x))2

d
d cos(x)
cot(x) =
= ( sin(x)·
dx
dx sin(x)
d
dx cot(x) =
f
g
You should now be able to find the derivative
Let’s use the quotient rule to find the derivative of the cotangent
function.
(x)
⇣ ⌘
(correct)
csc2 (x)
Answer(s) submitted:
• sin
• cos
• -1
Problem 11. (1 point)
Problem
(correct)
For f , g differentiable at x and g(x) 6= 0 we have
d
dx
d x+19
dx x+144 = (
✓ ◆
f
g(x) f 0 (x) f (x)g0 (x)
(x) =
g
g2 (x)
) /(x + 144)2
What are the values of x for which the derivative is not defined?
Answer(s) submitted:
• 125
• -144
(correct)
4
Problem 12. (1 point)
Problem
Problem 13. (1 point)
Problem
Suppose h is a function which has a derivative at x and f is the
function so that f (x) = 1/h(x) for all numbers x is the domain of
h so h(x) 6= 0.
Find the derivative of y = (2x2 + 1)27
(a) Find a function g so that f (x) = g(h(x)) for all numbers xin
the domain of f :
Answer(s) submitted:
• 2xˆ2+1
• 4x
y0 (x) = 27(
g(y) =
)27 1 ⇥
(correct)
Problem 14. (1 point)
Problem
(b) Use the chain rule to find f 0 (x).
Enter 99 when done.
Answer(s) submitted:
• 1/y
• 99
Find the derivative of y = [u(x)]a , supposing that u0 (x) exists.
y0 =
(correct)
[u(x)]a 1 ⇥
Answer(s) submitted:
• a
• u
(correct)
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