﻿ Derivatives of Second, Higher Order Derivatives Worksheet | Problems & Solutions

# Derivatives of Second, Higher Order Derivatives Worksheet

Derivatives of Second, Higher Order Derivatives Worksheet
• Page 1
1.
Find ′′($t$) to the function ($t$) = ln (10 + $t$2).
 a. 2$t$ b. - $\frac{1}{{\left(10+{t}^{2}\right)}^{2}}$ c. $\frac{2\left(10-{t}^{2}\right)}{{\left(10+{t}^{2}\right)}^{2}}$ d. $\frac{2\left(10+3{t}^{2}\right)}{10+{t}^{2}}$ e. $\frac{2\left(10+3{t}^{2}\right)}{{\left(10+{t}^{2}\right)}^{2}}$

#### Solution:

(t) = ln (10 + t2)
[Write the function.]

′(t) = ddt [ln (10 + t2)]
[Find ′(t).]

= 110+t2 (2t)
[Use Chain Rule.]

= 2t10+t2
[Simplify.]

′′(t) = ddt [2t10+t2]
[Find ′′(t).]

= (10+t2)(2)-2t(2t)(10+t2)2
[Use Quotient Rule.]

= 20+2t2-4t2(10+t2)2

= 2(10-t2)(10+t2)2
[Simplify.]

Therefore, ′′(t) = 2(10-t2)(10+t2)2

2.
Find $f$ ′′($x$) to the function $f$($x$) = 20 ${\left({x}^{2}+2\right)}^{\frac{5}{2}}$.

 a. IV b. I c. V d. III e. II

#### Solution:

f(x) = 20(x2+2)52
[Write the function.]

f ′(x) = ddx [20(x2+2)52]
[Find f ′(x).]

= 20(5 / 2)(x2 + 2)52-1 (2x)
[Use Power Rule, Chain Rule.]

= 100x(x2+2)32
[Simplify.]

f ′′(x) = ddx [100x(x2+2)32]
[Find f ′′(x).]

= 100(x)(32)(x2+2)32-1(2x) + 100(x2+2)32(1)
[Use Product Rule.]

= 300x2(x2+2)12 + 100(x2+2)32
[Simplify.]

= 200(x2+2)12(2x2 + 1)

Therefore, f ′′(x) = 200(x2+2)12(2x2 + 1).

3.
Find $g$ ′′($w$) to the function $g$($w$) = ${e}^{1-2{w}^{3}}$ + sin $w$.

 a. III b. IV c. V d. II e. I

#### Solution:

g(w) = e1-2w3 + sin w
[Write the function.]

g ′(w) = ddw(e1-2w3 + sin w)
[Find g ′(w).]

= e1-2w3(- 6w2) + cos w
[Use Chain Rule, Sum Rule.]

= - 6w2 e1-2w3 + cos w
[Simplify.]

g ′′(w) = ddw(- 6w2 e1-2w3 + cos w)
[Find g ′′(w).]

= - 6w2e1-2w3(- 6w2) + e1-2w3(- 12w) - sin w
[Use Chain Rule.]

= 36w4e1-2w3 - 12we1-2w3 - sin w
[Simplify.]

Therefore, g ′′(w) = 36w4e1-2w3 - 12we1-2w3 - sin w

4.
If $f$($t$) = 3${t}^{4}+8{t}^{\frac{1}{2}}+{e}^{2t}$, then find $f$ ′′′($t$).

 a. II b. V c. III d. IV e. I

#### Solution:

f(t) = 3t4+8t12+e2t
[Write the function.]

f ′ (t) = ddt(3t4+8t12+e2t)
[Find f ′(t).]

= 12t3 + 8(1 / 2t-12) + e2t(2)
[Use Sum Rule.]

= 12t3+4t-12+2e2t
[Simplify.]

f ′′(t) = ddt(12t3+4t-12+2e2t)
[Find f ′′(t).]

= 36t2 + 4(- 1 / 2t-12-1) + 2e2t(2)
[Use Sum Rule.]

= 36t2-2t-32+4e2t
[Simplify.]

f ′′′(t) = ddt(36t2-2t-32+4e2t)
[Find f ′′′(t).]

= 72t - 2(- 3 / 2t-32-1) + 4e2t(2)
[Use Difference, Sum Rule.]

= 72t + 3t-52+8e2t
[Simplify.]

Therefore, f ′′′(t) = 72t + 3t-52+8e2t

5.
Find $g$ ′′($r$) to the function $g$($r$) = sin 7$r$ + 3$r$4 + $e$- 8$r$.
 a. - 49sin 7$r$ + ${r}^{2}+{e}^{-8r}$ b. - 49sin 7$r$ + 36$r$2 + 64$e$- 8$r$ c. sin $r$ + 36${r}^{2}+{e}^{-8r}$ d. - 49sin $r$ - 36${r}^{2}+{e}^{-8r}$ e. 49sin 7$r$ + 36${r}^{2}-{e}^{-8r}$

#### Solution:

g(r) = sin 7r + 3r4 + e- 8r
[Write the function.]

g ′ (r) = ddr(sin 7r + 3r4 + e- 8r)
[Find g ′(r).]

= 7cos 7r + 12r3 + e- 8r(- 8)
[Use Sum Rule.]

= 7cos 7r + 12r3 - 8e- 8r
[Simplify.]

g ′′(r) = ddr(7cos 7r + 12r3 - 8e- 8r)
[Find g ′′(r).]

= - 49sin 7r + 36r2 - 8e- 8r(- 8)
[Use Sum Rule, Difference Rule.]

= - 49sin 7r + 36r2 + 64e- 8r
[Simplify.]

Therefore, g ′′ (r) = - 49sin 7r + 36r2 + 64e- 8r

6.
If $y$ = 10sin 4$x$ + $x$4, then find $\frac{{d}^{4}y}{d{x}^{4}}$.
 a. 112sin 4$x$ - 24 b. 112sin 2$x$ + 24 c. 256sin 4$x$ + 24 d. 112sin 4$x$ + 1 e. 10sin $x$ + 24

#### Solution:

y = 7sin 2x + x4
[Write the function.]

dydx = ddx(7sin 2x + x4)
[Find dydx.]

= 7(cos 2x)(2) + 4x3
[Use Sum Rule.]

= 14cos 2x + 4x3

d2ydx2 = ddx(14cos 2x + 4x3)
[Find d2ydx2.]

= 14(- sin 2x)(2) + 12x2
[Use Sum Rule.]

= - 28sin 2x + 12x2
[Simplify.]

d3ydx3 = ddx(- 28sin 2x + 12x2)
[Find d3ydx3.]

= - 28(cos 2x)(2) + 24x
[Use Sum Rule.]

= - 56cos 2x + 24x
[Simplify.]

d4ydx4 = ddx(- 56cos 2x + 24x)
[Find d4ydx4.]

= - 56(- sin 2x)(2) + 24
[Use Sum Rule.]

= 112sin 2x + 24
[Simplify.]

Therefore, d4ydx4 = 112sin 2x + 24

7.
Find $g$ ′′′($s$) to the function $g$($s$) = sin 5$s$ + $e$- 3$s$ + ln 5$s$.
 a. - 125cos 5$s$ - 27${e}^{-3s}+\frac{2}{{s}^{3}}$ b. - 125cos 5$s$ - ${e}^{-3s}+\frac{2}{{s}^{3}}$ c. - 125cos 5$s$ - 27${e}^{-3s}+\frac{6}{{5s}^{3}}$ d. - 5cos 5$s$ - 27${e}^{-3s}+\frac{2}{{s}^{3}}$ e. - 125cos 5$s$ - ${e}^{-3s}-\frac{2}{{s}^{3}}$

#### Solution:

g(s) = sin 5s + e- 3s + ln 5s
[Write the function.]

g ′(s) = dds(sin 5s + e- 3s + ln 5s)
[Find g ′(s).]

= (cos 5s)(5) + e- 3s(- 3 ) + 15s(5)
[Use Sum Rule.]

= 5cos 5s - 3e- 3s + 1s
[Simplify.]

g ′′(s) = dds(5cos 5s - 3e- 3s + 1s)
[Find g ′′(s).]

= 5(- sin 5s)(5) - 3e- 3s(- 3) - 1s2
[Use Sum Rule, Difference Rule.]

= - 25sin 5s + 9e- 3s - 1s2
[Simplify.]

g ′′′(s) = dds(- 25sin 5s + 9e- 3s - 1s2)
[Find g ′′′(s).]

= - 25(cos 5s)(5) + 9e- 3s(- 3) - (- 2)1s3
[Use Sum Rule, Difference Rule.]

= - 125cos 5s - 27e- 3s + 2s3
[Simplify.]

Therefore, g ′′′ (s) = - 125cos 5s - 27e- 3s + 2s3

8.
Find ′′′($q$) to the function ($q$) = ln $q$ + 7$q$2 + sin 9$q$.
 a. $\frac{1}{{q}^{3}}$ - 729cos 9$q$ b. $\frac{2}{{q}^{3}}$ - 729cos 9$q$ c. - 729cos 9$q$ d. $\frac{2}{{q}^{3}}$ + 729cos 9$q$ e. $\frac{2}{{q}^{3}}$ + 14 - 729cos 9$q$

#### Solution:

(q) = ln q + 7q2 + sin 9q
[Write the function.]

′ (q) = ddq(ln q + 7q2 + sin 9q)
[Find ′ (q).]

= 1q + 14q + 9cos 9q
[Use Sum Rule.]

′′ (q) = ddq(1q + 14q + 9cos 9q)
[Find ′′ (q).]

= - 1q2 + 14 - 81sin 9q
[Use Sum Rule.]

′′′(q) = ddq(- 1q2 + 14 - 81sin 9q)
[Find ′′′(q).]

= 2q3 - 729cos 9q
[Use Sum Rule, Difference Rule.]

Therefore, ′′′(q) = 2q3 - 729cos 9q

9.
Find ′′($y$) to the function ($y$) = 14${y}^{\frac{5}{2}}$ + tan 4$y$.
 a. ${y}^{\frac{1}{2}}$ + 32sec 4$y$ b. ${y}^{\frac{1}{2}}$ + sec2 4$y$ tan 4$y$ c. $\frac{105}{2}$${y}^{\frac{1}{2}}$ - sec 4$y$ tan 4$y$ d. ${y}^{\frac{1}{2}}$ - 32sec 4$y$ e. $\frac{105}{2}$${y}^{\frac{1}{2}}$ + 32sec24$y$ tan 4$y$

#### Solution:

(y) = 14y52 + tan 4y
[Write the function.]

′(y) = ddy(14y52 + tan 4y)
[Find ′(y).]

= 14(5 / 2)y52-1 + 4sec2 4y
[Use Sum Rule.]

= 35y32 + 4sec2 4y
[Simplify.]

′′(y) = ddy(35y32 + 4sec2 4y)
[Find ′′(y).]

= 35(3 / 2)y32-1 + 8(sec 4y)(sec 4y tan 4y)(4)
[Use Sum Rule.]

= 105 / 2y12 + 32sec2 4y tan 4y

Therefore, ′′(y) = 105 / 2y12 + 32sec2 4y tan 4y

10.
If $g$($r$) = ${r}^{3}+4{r}^{2}$ - 5$r$ + 9, then find $g$ ′′($r$) at $r$ = 2.
 a. 6 b. 14 c. 20 d. 784 e. 122

#### Solution:

g(r) = r3+4r2 - 5r + 9
[Write the function.]

g ′ (r) = ddr(r3+4r2 - 5r + 9)
[Find g ′(r).]

= 3r2 + 8r - 5
[Use Sum Rule, Difference Rule.]

g ′′(r) = ddr(3r2 + 8r - 5)
[Find g ′′(r).]

= 6r + 8
[Use Sum Rule, Difference Rule.]

g ′′(2) = 6(2) + 8
[Find g ′′(2).]

= 12 + 8 = 20
[Simplify.]

Therefore, g ′′(r) at r = 2 is 20.