Calculus I - Calculus Detective #2 - Solutions
Spring 2007
Pat Rossi & Sraboni Ghosh
Name
Instructions. Each problem below is accompanied by a solution. Each solution contains
at least one serious error, but no more than two serious errors. 1 Explain what the error
is (i.e., explain what the person did wrong). 2 Explain what the person should have done.
3
Re-work the problem correctly.
1.
d
dx
d
dx
h
10
(x3 − 3x + 5)
h
10
(x3 − 3x + 5)
i
=
i
9
= 10 (x3 − 3x + 5)
Error: The function whose derivative we are computing is of the form (g (x))n .
By the Chain Rule,
d
dx
[(g (x))n ] = n (g (x))n−1 · g 0 (x)
The person computed n (g (x))n−1 , but didn’t multiply by the derivative, g 0 (x) .
This is what they should have done:
d
dx
2.
d
dx
d
dx
⎡
⎢³
⎢ 3
⎢ x
⎣|
h
⎤
´10 ⎥
⎥
⎥
}⎦
− 3x + 5
{z
(g(x))n
10
(x3 − 3x + 5)
h
10
(x3 − 3x + 5)
i
i
³
´9
= 10 x3 − 3x + 5
|
{z
n(g(x))n−1
}
³
´
· 3x2 − 3
|
{z
g 0 (x)
}
=
9
= 10 (3x2 − 3)
Error: The function whose derivative we are computing is of the form (g (x))n .
By the Chain Rule,
d
dx
[(g (x))n ] = n (g (x))n−1 · g 0 (x)
The person computed n (g0 (x))n−1 .
There are actually two mistakes here:
First, they didn’t multiply n ( )n−1 by g 0 (x) .
Second, they replaced the “inner function” g (x) with g 0 (x) . When we compute a
derivative using the chain rule, we don’t compute the derivative of the “inner function”
until it’s time to multiply n ( )n−1 by the derivative of the “inner function” g 0 (x) .
This is what they should have done:
d
dx
3.
⎡
⎢³
⎢ 3
⎢ x
⎣|
´10 ⎥
⎥
⎥
}⎦
− 3x + 5
{z
(g(x))n
⎤
³
´9
= 10 x3 − 3x + 5
|
{z
n(g(x))n−1
d
dx
[sin (2x3 + 8x2 )] =
d
dx
[sin (2x3 + 8x2 )] = cos (6x2 + 16x)
}
³
´
· 3x2 − 3
|
{z
g 0 (x)
}
Error: The function whose derivative we are computing is of the form f (g (x)) .
⎡
By the Chain Rule,
⎢
⎢
⎢
d ⎢
dx ⎢
⎢
⎣
⎤
f (g (x))
↑
outer inner
⎥
⎥
⎥
⎥
⎥
⎥
⎦
f 0 (g (x))
=
|
{z
}
Deriv. of outer
eval. at inner
·
g0 (x)
| {z }
Deriv. of inner
The person computed f 0 (g 0 (x)) , so again, there are actually two mistakes here.
First, they didn’t multiply f 0 ( ) by the derivative of the “inner function” g 0 (x) .
Second, they replaced the “inner function” g (x) with g 0 (x) . When we compute a
derivative using the chain rule, we don’t compute the derivative of the “inner function”
until it’s time to multiply f 0 (g (x)) by the derivative of the “inner function” g 0 (x) .
This is what they should have done:
d
dx
⎡
⎤
⎢
³
´⎥
⎢
⎢|{z}
⎥
sin 2x3 + 8x2 ⎥
⎣
{z
}⎦
|
f
(g(x))
³
´
³
´
= cos 2x3 + 8x2 · 6x2 + 16x
|
{z
f 0 (g(x))
}
{z
|
g 0 (x)
}
Alternatively, in terms of the “outer” and “inner” functions, we can write:
⎡
⎢
⎢
⎢
d ⎢
sin (2x3
dx ⎢
⎢
⎣ ↑
⎤
⎥
⎥
⎥
2 ⎥
+ 8x )⎥
⎥
↑
⎦
outer inner
³
= cos 2x3 + 8x2
|
{z
´
}
·
Deriv. of outer
eval. at inner
³
|
´
6x2 + 16x
{z
}
Deriv. of inner
2
4.
d
dx
[sin (cos (x))] =
d
dx
⎡
⎤
⎢
sin (cos (x))⎥
⎣|{z}
⎦
| {z }
1st
= cos (x) cos (x) + (− sin (x))sin (x) = cos2 (x) − sin2 (x)
| {z } | {z }
1st prime
2n d
|
2n d
{z
2n d prime
}| {z }
1st
Error: This person used the product rule to compute the derivative, but sin (cos (x))
is not a product.
sin (cos (x)) is the function sin ( ) evaluated at cos (x) . (i.e. we get sin (cos (x)) by
“plugging cos (x) into” sin ( ) .)
Remark: For the sake of comparison, note that sin (x) cos (x) is a product, but that
sin (cos (x)) is not a product.
This is what they should have done:
d
dx
⎡
⎤
⎢
⎥
⎢ sin (cos (x))⎥
⎣|{z}| {z }⎦
f
= cos (cos (x)) · (− sin (x)) = − cos (cos (x)) · sin (x)
|
(g(x))
{z
f 0 (g(x))
}
|
{z
g 0 (x)
}
Alternatively, in terms of the “outer” and “inner” functions, we can write:
⎡
⎢
⎢
⎢
d ⎢
dx ⎢
⎢
⎣
⎤
sin (cos (x))
↑
↑
outer inner
⎥
⎥
⎥
⎥
⎥
⎥
⎦
=
cos (cos (x))
|
{z
}
·
Deriv. of outer
eval. at inner
(− sin (x))
|
{z
}
= − cos (cos (x)) · sin (x)
Deriv. of inner
5. The equation below defines y implicitly as a function of x. Compute y 0 (i.e., compute
dy
)
dx
3x2 + x2 y 3 = sin (y)
1. Differentiate both sides w.r.t. x
d
dx
[3x2 + x2 y 3 ] =
d
dx
[sin (y)]
⇒ 6x + (2x) (3y 2 · y 0 ) = cos (y) · y 0
⇒ 6x + 6xy 2 · y 0 = cos (y) · y 0
2. Get y 0 terms on the L.H. side and all other terms on the R.H. side.
⇒ 6xy 2 · y 0 − cos (y) · y 0 = −6x
3. Factor out y 0
⇒ y 0 [6xy 2 − cos (y)] = −6x
4. Divide both sides by the “co-factor” of y 0 .
6x
⇒ y 0 = − 6xy2 −cos(y)
3
Error: In the first step, when they computed the derivative of x2 y 3 (w.r.t. x), they
neglected to consider the fact that x2 y 3 is a product. Thus, they should have used the
product rule:
d
dx
[x2 y 3 ] = 2xy 3 + 3y 2 y 0
This is how the problem should have been solved:
1. Differentiate both sides w.r.t. x
d
dx
[3x2 + x2 y 3 ] =
d
dx
[sin (y)]
⇒ 6x+ 2xy 3 + 3y 2 y 0 = cos (y) · y 0
|
{z
}
by the Product Rule
2. Get y 0 terms on the L.H. side and all other terms on the R.H. side.
⇒ 3y 2 y 0 − cos (y) · y 0 = −6x − 2xy 3
3. Factor out y 0
⇒ y 0 [3y 2 − cos (y)] = −6x − 2xy 3
4. Divide both sides by the “co-factor” of y 0 .
3
⇒ y 0 = − 3y6x+2xy
2 −cos(y)
6.
d
dx
d
dx
∙³
∙³
4x3 +5x−3
8x2 +5x+5
4x3 +5x−3
8x2 +5x+5
´10 ¸
´10 ¸
=
= 10
³
4x3 +5x−3
8x2 +5x+5
´9
·
12x2 +5
16x+5
Error: The person correctly understood that they were supposed to use the chain rule
d
[(g (s)n )] = n (g (x))n−1 · g0 (x):
dx
⎡
⎢Ã
3
d ⎢
⎢ 4x + 5x − 3
⎢
2
dx ⎢
⎣ 8x + 5x + 5
|
{z
(g(x))n
⎤
!10 ⎥
⎥
⎥
⎥
⎥
}⎦
Ã
4x3 + 5x − 3
=10
8x2 + 5x + 5
|
{z
n(g(x))n−1
!9
}
·
12x2 + 5
16x + 5
| {z }
g 0 (x)
Their mistake was that in applying the chain rule, they neglected to consider the fact
that the “inner function” g (x) is a quotient. Thus, they should have used the quotient
rule when computing the derivative of the “inner function” g (x) .
This is what they should have done:
⎡
⎢Ã
3
d ⎢
⎢ 4x + 5x − 3
⎢
2
dx ⎢
⎣ 8x + 5x + 5
|
{z
(g(x))n
⎤
!10 ⎥
⎥
⎥
⎥
⎥
}⎦
Ã
4x3 + 5x − 3
= 10
8x2 + 5x + 5
|
{z
n(g(x))n−1
!9
}
(12x2 + 5) (8x2 + 5x + 5) − (16x + 5) (4x3 + 5x − 3)
·
[8x2 + 5x + 5]2
|
4
{z
g 0 (x)
}