MATH 100 V1A
September 26th – Practice problems
Solutions
√
1. Let f (x) =
x
.
x+1
What is f 0 (2)?
Solution: Using the Quotient Rule we get that
f 0 (x) =
1
√
2 x
· (x + 1) −
(x +
√
x·1
.
1)2
Note that you do not need to simplify your answer unless you are explicitly told to do
so.
2. Differentiate f (x) =
x
.
1
2x+ 3x+2
Solution: Again, using the Quotient Rule (twice) we have that
1
0−1·3
1 · 2x + 3x+2 + x · 2 + (3x+2)2
f 0 (x) =
2
1
2x + 3x+2
1
3
2x + 3x+2
+ x · 2 − (3x+2)
2
=
2
1
2x + 3x+2
3. Find the equation of the line tangent to y =
2√
3−4 x
at y = −2.
Solution: Let f (x) = 3−42√x . Let’s first solve for the x-value that corresponds to the
√
√
y = −2. We have that −2 = 3−42√x if and only if −(3 − 4 x) = 1, or 4 x = 4, which
has solutions x = ±1. Clearly, x = −1 is not in the domain of f , so the solution we
want is x = 1.
Now, by the Quotient Rule,
0 − 2 · 0 − 4 · 2√1 x
4
√ 2
√ .
f 0 (x) =
=√
(3 − 4 x)
x(3 − 4 x)2
So, the slope of the tangent line is given by f (1) =
tangent line is therefore given by
y − (−2) = 4(x − 1)
or
4
1·(3−4)2
= 4. The equation of the
y = 4x − 6.
4. Make a conjecture as to what is the nth derivative of
this conjecture?
Solution: Let f (x) =
Rule) we get
1
.
x−1
1
.
x−1
What is required to prove
If we calculate the first few derivatives (using the Quotient
f 0 (x) =
0−1
1
=
−
(x − 1)2
(x − 1)2
f 00 (x) = −
f 000 (x) =
0 − 2(x − 1)
2
=
4
(x − 1)
(x − 1)3
0 − 2 · 3(x − 1)2
6
=−
6
(x − 1)
(x − 1)4
f (4) (x) = −
24
0 − 6 · 4(x − 1)3
=
,
8
(x − 1)
(x − 1)5
so we might conjecture that f (n) (x) = (−1)n (x−1)n!(n+1) , where f (n) is the nth derivative
of f , and n! = 1 · 2 · · · (n − 1) · n (read n factorial). This is in fact the correct formula,
and to prove it we will use mathematical induction. We have already established that
the formula is true when n = 1, 2, 3, 4. Now assume that the formula is true when
n = k for some positive integer k. We need to show that this implies that it is also
true for n = k + 1. The formula when n = k is:
f (k) (x) = (−1)k
k!
.
(x − 1)(k+1)
To get the (k + 1)th derivative of f (x), we can just take the derivative of f (k) (x) using
the above formula. That is
d (k)
d
k!
(k+1)
k
f
(x) =
f (x) =
(−1)
dx
dx
(x − 1)(k+1)
= (−1)k · k! · −(k + 1)(x − 1)−(k+1)−1
k+1
k
= −(−1) · k! ·
(x − 1)(k+2)
(k + 1)!
= (−1)(k+1)
,
(x − 1)(k+2)
which is just the formula when n = k +1. So we know that it is true when n = 1, 2, 3, 4,
and that if it is true when n = k then it must also be true when n = k + 1. This means
that we can conclude that it must also be true for n = 5, n = 6 and, in fact, that it
must be true for any positive integer n.
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