# calculus project

Marquise Andrews
Math 412
November 13, 2019
1.
A. Leibniz’s Rule is used to find higher order derivatives, typically expressed to the n-th
order. It is used when there two components that need to be derived simultaneously. Since it can
be tedious to do it long-hand, Leibniz’s Rule is utilized when finding derivatives of orders higher
than five. The following function can be given to find derivatives to the n-th order.
(𝑢𝑣)𝑛 ∑𝑛𝑖=1(𝑛𝑖)𝑢𝑛−𝑖 𝑣 𝑖 , where (𝑛𝑖) is the amount of i-combinations of n elements
(Math24. (2019). Leibniz Formula. [online] Available at:
https://www.math24.net/leibniz-formula/ [Accessed 4 Nov. 2019].)
𝑑20
B. find 𝑑20 𝑥 (𝑥 2 𝑒 4𝑥 ).
u=𝑒 4𝑥
v=𝑥 2
n= 20
20
4𝑥 20−𝑖
(𝑥 2 𝑒 4𝑥 )20 ∑20
(𝑥 2 )𝑖 = (20
)(𝑒 4𝑥 )20−𝑖 (𝑥 2 )𝑖 =
𝑖=0( 𝑖 )(𝑒 )
𝑖
(20
)(𝑒 4𝑥 )20 (𝑥 2 )+(20
)(𝑒 4𝑥 )19 (𝑥 2 )′ +(20
)(𝑒 4𝑥 )18 (𝑥 2 )′′+(20
)(𝑒 4𝑥 )17 (𝑥 2 )′′′ +(20
)(𝑒 4𝑥 )16 (𝑥 2 )4 +
0
1
2
3
4
(20
)(𝑒 4𝑥 )15 (𝑥 2 )5 +(20
)(𝑒 4𝑥 )14 (𝑥 2 )6 +(20
)(𝑒 4𝑥 )13 (𝑥 2 )7 +(20
)(𝑒 4𝑥 )12 (𝑥 2 )8+(20
)(𝑒 4𝑥 )11 (𝑥 2 )9+
5
6
7
8
9
(20
)(𝑒 4𝑥 )10 (𝑥 2 )10 +(20
)(𝑒 4𝑥 )9 (𝑥 2 )11 +(20
)(𝑒 4𝑥 )8 (𝑥 2 )12+(20
)(𝑒 4𝑥 )7 (𝑥 2 )13+(20
)(𝑒 4𝑥 )6 (𝑥 2 )14+
10
11
12
13
14
(20
)(𝑒 4𝑥 )5 (𝑥 2 )15 +(20
)(𝑒 4𝑥 )4 (𝑥 2 )16 +(20
)(𝑒 4𝑥 )3 (𝑥 2 )17+(20
)(𝑒 4𝑥 )2 (𝑥 2 )18+(20
)(𝑒 4𝑥 )1 (𝑥 2 )19+
15
16
17
18
19
(20
)(𝑒 3𝑥 )(𝑥 2 )20
20
As you evaluate each piece, anything after the third term in the expansion will
yield 0 as 𝑥 derived twice is 2 and the third derivative will be 0 thereafter. Thus, we are only
looking at the first three terms.
2
(20
)(𝑒 4𝑥 )20 (𝑥 2 )+(20
)(𝑒 4𝑥 )19 (𝑥 2 )′ +(20
)(𝑒 4𝑥 )18 (𝑥 2 )′′ =
0
1
2
3386784401 𝑥 2 𝑒 4𝑥 + 12490458680 𝑥𝑒 3𝑥 + 217219785820 =
𝐴𝑛𝑠𝑤𝑒𝑟: 677420489 𝑒 4𝑥 (16𝑥 2 + 240𝑥 + 380)
2.
Find P14(-1):
1
1
the first part is a constant 214 14! = 1428329123020800
Marquise Andrews
Math 412
November 13, 2019
now we find the second part which can be determined using the binomial theorem
(𝑥 2 − 1)14 =&gt;(𝑥 2 − 1)2 (𝑥 2 − 1)13 =&gt; (𝑥 2 − 1)1 (𝑥 2 − 1)6 (𝑥 2 − 1)6 =&gt;
(𝑥 2 − 1)1 (𝑥 2 − 1)2 (𝑥 2 − 1)2 (𝑥 2 − 1)2 (𝑥 2 − 1)2 (𝑥 2 − 1)2 (𝑥 2 − 1)2 =&gt;
(𝑥 2 − 1)1 (𝑥 4 − 2𝑥 2 + 1)6 =&gt; (𝑥 2 − 1)1 (𝑥 4 − 2𝑥 2 + 1)2 (𝑥 4 − 2𝑥 2 + 1)2 (𝑥 4 − 2𝑥 2 + 1)2
=&gt; (𝑥 2 − 1)1 (𝑥 8 − 4𝑥 6 + 6𝑥 4 − 4𝑥 2 + 1)3
 (𝑥 28 + 14𝑥 26 + 91𝑥 24 + 364𝑥 22 + 1001𝑥 20 + 2002𝑥18 + 3003𝑥16 + 3432𝑥14 +
3003𝑥12 + 2002𝑥 10 + 1001𝑥 8 + 364𝑥 6 + 91𝑥 4 + 14𝑥 2 + 1).
So now we take the derivative of this product
1st : 28𝑥 27 + 364𝑥 25 + 4368𝑥 23 + 8008𝑥 21 + 20020𝑥19 + 36036𝑥17 + 48048𝑥15 +
48048𝑥13 + 36030𝑥11 + 20020𝑥 9 + 8008𝑥 7 + 2184𝑥 5 + 28𝑥).
2nd : 756𝑥 26 + 9100𝑥 24 + 100464𝑥 22 + 168168𝑥 20 + 380380𝑥18 + 612612𝑥16 +
720720𝑥14 + 624624𝑥12 + 396396𝑥10 + 180180𝑥 8 + 56056𝑥 6 + 10920𝑥 4 + 28).
3rd :19656𝑥 25 + 218400𝑥 23 + 2210208𝑥 21 + 363360𝑥19 + 6846840𝑥17 + 9801792𝑥15 +
100090080𝑥13 + 749588𝑥11 + 1441440𝑥 7 + 336336𝑥 5 + 43680𝑥 3
4th : 491400𝑥 24 + 502320𝑥 22 + 46414368𝑥 20 + 63903840𝑥18 + 115396280𝑥16 +
147026880𝑥14 + 131171040𝑥12 + 82450368𝑥10 + 35675640𝑥 8 + 10090080𝑥 6 +
1681680𝑥 4 + 131040𝑥 2
5th : 11793600𝑥 23 + 110510400𝑥 21 + 928287360𝑥19 + 1150269120𝑥17 +
1862340480𝑥15 + 205876320𝑥 13 + 1574052480𝑥11 + 82450380𝑥 9 + 285405120𝑥 7 +
60540480𝑥 5 + 6726720𝑥 3 + 262080𝑥
.
.
.
14th : 6900675250448000𝑥14 + 16838835658444800𝑥13 + 69028767295488000𝑥 9 −
137633578258176000𝑥 7 + 38147476663296000𝑥 5 + 4487398430976000𝑥 3 +
149597947699200𝑥
1
Now we apply 1428329123020800 and sub -1 in for x. When -1 is raised to an odd power, the
coefficient is negated
Marquise Andrews
1
510117543936
Math 412
November 13, 2019
(9600675250448000𝑥14 + 61838835658444800𝑥12 +
96028767295488000𝑥10 + 317633578258176000𝑥 8 + 38147476663296000𝑥 6 −
1
4487398430976000𝑥 4 + 149597947699200𝑥)= 510117543936 (8644793334e16)=
78956.06792
1
Find P13(1)= 1428329123020800 (9600675250448000𝑥14 + 61838835658444800𝑥12 +
96028767295488000𝑥10 + 317633578258176000𝑥 8 + 38147476663296000𝑥 6 +
1
4487398430976000𝑥 4 + 149597947699200𝑥)= 1428329123020800 (-8644793334e16)=
-78956.06792