Handout 5, Summer 2014
Math 1823-171
29 May 2014
1. Consider the following table of values:
x f (x) g(x) f 0 (x) g 0 (x)
3
4
8
1
11
4
3
4
2
9
8
8
3
9
4
Let h(x) = (f ◦ g)(x) and l(x) = g(f (x)). Compute h0 (3), h0 (4),
l0 (8), and l0 (3).
Solution. Since h(x) = f (g(x)), by the chain rule we have that h0 (x) =
f 0 (g(x)) · g 0 (x), so that we have:
h0 (3) = f 0 (g(3))g 0 (3) = f 0 (8) · 11 = 9 · 11 = 99
and
h0 (4) = f 0 (g(4))g 0 (4) = f 0 (4) · 9 = 2 · 9 = 18
Next, since l(x) = g(f (x)), we again have by the chain rule that
l0 (x) = g 0 (f (x))f 0 (x). Therefore,
l0 (8) = g 0 (f (8))f 0 (8) = g 0 (8) · 9 = 4 · 9 = 36
and
l0 (3) = g 0 (f (3))f 0 (3) = g 0 (4) · 1 = 9
We have the result.
2. Suppose g(x) =
g 0 (3).
p
f (x). Given the graph of f below, calculate
1
2
Solution. Note that g(x) is the composition of the functions o(x) =
and i(x) = f (x). Therefore, using the chain rule, we have
√
o(x) = x
√
x
i(x) = f (x)
1
o0 (x) = x−1/2
2
0
i (x) = f 0 (x)
Thus, g 0 (x) = (o(i(x)))0 = o0 (i(x))i0 (x) = 12 (f (x))−1/2 f 0 (x). This gives
f 0 (3)
. From the graph, we can immediately see that f (3) = 2.
g 0 (3) = √
2
f (3)
Furthermore, since the derivative of the graph at a point is the slope of
the line tangent to the graph at that point, we know that f 0 (x) = −2
.
3
−2
Indeed, the slope of the blue line in the graph is 3 . Therefore, we
have:
−2
1
f 0 (3)
−1
=
· √ = √
g 0 (3) = p
3 2 2
3 2
2 f (3)
This gives the result.
3. If g is twice differentiable (i.e. has a first and second derivative)
and f (x) = xg(x2 ), find f 00 (x).
Solution. Here we have a product of two functions: x and g(x2 ). Therefore, we use the product rule first to take the derivative of f :
f 0 (x) = x0 g(x2 ) + x(g(x2 ))0
= g(x2 ) + x(g(x2 ))0
Now, g(x2 ) is a composition of functions:
o(x) = g(x)
i(x) = x2
o0 (x) = g 0 (x)
i0 (x) = 2x
Therefore, using the chain rule, we have that (g(x2 ))0 = o0 (i(x))i0 (x) =
2xg 0 (x2 ). Thus, we have
f 0 (x) = g(x2 ) + 2x2 g 0 (x2 )
To find f 00 (x), we follow a similar procedure. We have:
f 00 (x) = 2xg 0 (x2 ) + 4xg 0 (x2 ) + 4x3 g 00 (x2 )
This gives the result.
3
103
d
4. Find dx
103 cos(2x) by computing the first few derivatives and
observing a pattern.
Solution. Let’s begin with the first derivative of cos(2x). This is a
composition of functions:
o(x) = cos(x)
i(x) = 2x
o0 (x) = − sin(x)
i0 (x) = 2
By the chain rule, then, we have that (cos(2x))0 = −2 sin(2x). Similarly, we can compute the next few derivatives:
(cos(2x))0 = −2 sin(2x)
(cos(2x))00 = −4 cos(2x)
(cos(2x))000 = 8 sin(2x)
(cos(2x))(4) = 16 cos(2x)
And so on. Note that every fourth derivative is some multiple of
cos(2x). Furthermore, the coefficient of the derivative is ± a power of
2. Putting this together, we can see that (cos(2x))(100) = 2100 cos(2x).
Therefore, we have:
(cos(2x))(101) = −2101 sin(2x)
(cos(2x))(102) = −2102 cos(2x)
(cos(2x))(103) = 2103 sin(2x)
We have the result.
5. Calculate
dy
dx
√
√
if 2 x + y = 3.
Solution. Replacing y with f (x), we have:
p
√
2 x + f (x) = 3
Differentiating both sides with respect to x, we obtain:
1 −1/2
1
2
x
+ f (x)−1/2 f 0 (x) = 0
2
2
That is,
1
f 0 (x)
√ + p
=0
x 2 f (x)
4
We now solve for f 0 (x). We have:
1
f 0 (x)
√ + p
=0
x 2 f (x)
f 0 (x)
−1
⇒ p
=√
x
2 f (x)
p
−2 f (x)
√
⇒ f 0 (x) =
x
Finally, swapping all f (x) for y and all f 0 (x) for y 0 , we arrive at:
r
√
−2 y
y
dy
0
=y = √
= −2
dx
x
x
6. Calculate
dy
dx
if 2x3 + x2 y − xy 3 = 2.
Solution. Here we will proceed without swapping y for f (x). See if you
can follow along. We differentiate both sides with respect to x. Note
that a product rule is needed on the second term of the left side, and
a product and chain rule are needed on the third term of the left side.
We have:
6x2 + (2xy + x2 y 0 ) − (y 3 + 3xy 2 y 0 ) = 0
⇒ 6x2 + 2xy + x2 y 0 − y 3 − 3xy 2 y 0 = 0
⇒ 6x2 + 2xy − y 3 = 3xy 2 y 0 − x2 y 0
⇒ 6x2 + 2xy − y 3 = y 0 (3xy 2 − x2 )
⇒ y0 =
dy
6x2 + 2xy − y 3
=
dx
3xy 2 − x2
We have the result.
7. Calculate
dy
dx
if 4 cos(x) sin(y) = 1.
Solution. Here we go into more explicit detail. First, let’s swap y for
f (x). Then we have
4 cos(x) sin(f (x)) = 1
To differentiate both sides, we must use a product and chain rule on
the left-hand side. That is, by the product rule we have:
4(− sin(x)) sin(f (x)) + 4 cos(x)(sin(f (x)))0 = 0
5
Finding (sin(f (x)))0 and plugging it into this equation will complete
the differentiation. To compute this, we must use the chain rule.
o(x) = sin(x)
i(x) = f (x)
o0 (x) = cos(x)
i0 (x) = f 0 (x)
Therefore, by the chain rule (sin(f (x)))0 = cos(f (x))f 0 (x). Plugging
this into the equation above, we have
−4 sin(x) sin(f (x)) + 4 cos(x) cos(f (x))f 0 (x) = 0
At last, solving for f 0 (x), we have:
f 0 (x) =
4 sin(x) sin(f (x))
4 cos(x) cos(f (x))
Therefore, swapping f (x) back for y and f 0 (x) back for y 0 , we arrive at
our final answer:
sin(x) sin(y)
dy
=
y0 =
dx
cos(x) cos(y)
8. Calculate
dy
dx
if
√
x + y = 1 + x2 y 2 .
Solution. Again, we will work out this result in detail. We begin by
replacing y by f (x). We have:
p
x + f (x) = 1 + x2 f (x)2
dy
To find dx
, we need to start by differentiating both sides. To accomplish
this, we must use a chain rule on the left-hand side, and a product and
chain rule on the right-hand side. For the left-hand side, we have:
√
o(x) = x
i(x) = x + f (x)
1
o0 (x) = x−1/2
2
0
i (x) = 1 + f 0 (x)
p
0
Therefore, we have that
x + f (x) = 12 (x + f (x))−1/2 (1 + f 0 (x)).
For the right-hand side, we must use the sum and product rules first,
giving:
d
1 + x2 f (x)2 = 2xf (x)2 + x2 (f (x)2 )0
dx
6
Again, to compute (f (x)2 )0 , we must use the chain rule:
o(x) = x2
i(x) = f (x)
o0 (x) = 2x
i0 (x) = f 0 (x)
Therefore, (f (x)2 )0 = 2f (x)f 0 (x) by the chain rule. Therefore, we see
that the derivative of the equation at the top of the problem is:
1 + f 0 (x)
p
= 2xf (x)2 + 2x2 f (x)f 0 (x)
2 x + f (x)
Now we need to solve for f 0 (x), and we will be finished. We have:
1 + f 0 (x)
p
= 2xf (x)2 + 2x2 f (x)f 0 (x)
2 x + f (x)
p
⇒ 1 + f 0 (x) = 2xf (x)2 + 2x2 f (x)f 0 (x) 2 x + f (x)
p
p
2
0
2
0
⇒ f (x) − 2x f (x)f (x) 2 x + f (x) = 2xf (x) 2 x + f (x) − 1
p
2xf (x)2 2 x + f (x) − 1
p
⇒ f 0 (x) =
1 − 2x2 f (x) 2 x + f (x)
dy
for f 0 (x) and y for f (x) gives the result, after a little
Swapping dx
cleaning up:
√
4xy 2 x + y − 1
dy
√
=
dx
1 − 4x2 y x + y
9. Find the equation of the line tangent to the curve x2 +xy +y 2 =
3 at the point (1, 1).
Solution. In order to find the equation of the tangent line at (1, 1),
we need the slope of the line and a point on the line. Of course, we
know immediately that (1, 1) is a point on the line. Recall that the
dy
slope of the tangent line is given by dx
evaluated at (x, y) = (1, 1).
7
Differentiating, we have
d
x2 + xy + y 2 = 3
dx
= 2x + (y + xy 0 ) + 2yy 0 = 0
⇒ xy 0 + 2yy 0 = −2x − y
⇒ y 0 (x + 2y) = −2x − y
−2x − y
⇒ y0 =
x + 2y
(If you don’t see how the differentiation worked, write it out for yourself, using the worked-out examples above as a guide.) Evaluating at
(x, y) = (1, 1), we have that the slope of the tangent line is −2−1
= −1.
1+2
Therefore, We have that the tangent line is
y − 1 = −1(x − 1)
Or, in slope-intercept form:
y =2−x
This concludes the result.