﻿ Derivatives of Logarithmic Functions Worksheet | Problems & Solutions

# Derivatives of Logarithmic Functions Worksheet

Derivatives of Logarithmic Functions Worksheet
• Page 1
1.
If $g$($l$) = ln (sec 52$l$ + tan 52$l$), then find $g$′ ($l$).
 a. b. 52sec 52$l$ c. d. sec 52$l$ e.

#### Solution:

g(l) = ln (sec 52l + tan 52l)
[Write the function.]

g′ (l) = ddl( ln (sec 52l + tan 52l))
[Find g′ (l)]

= 1sec 52l+tan 52lddl(sec 52l+tan 52l)
[Use the Chain Rule.]

= 52sec 52ltan 52l+52sec252lsec 52l+tan 52l

= 52sec 52l(tan 52l+sec 52l)sec 52l+tan 52l = 52sec 52l
[Factor out 52sec 52l and simplify.]

g′ (l) = 52sec 52l

2.
If $g$($q$) = $q$10ln 8$q$, then find $g$′ ($q$).
 a. b. 10 c. d. e.

#### Solution:

g(q) = q10ln 8q
[Write the function.]

g ′(q) = ddq(q10ln 8q)
[Find g ′(q).]

= q10ddq(ln 8q)+(ln 8q)ddq(q10)
[Use the Product Rule.]

= q10(1q)+(ln 8q)(10q9)
[Use the Chain Rule.]

= q9(1+10ln 8q)

g ′(q) = q9(1+10ln 8q)

3.
If ($x$) = $x$(ln 8$x$)11, then find ′ ($x$).
 a. (ln 8$x$)10($\frac{11}{8}$ + (ln 8$x$)) b. (ln 8$x$)10($\frac{11}{8}$ - (ln 8$x$)) c. (ln 8$x$)10(11 + (ln 8$x$)) d. (10(ln 8$x$))(11 + (ln 8$x$)) e. (ln 8$x$)10(11 - (ln 8$x$))

#### Solution:

(x) = x(ln 8x)11
[Write the function.]

′ (x) = ddx(x(ln 8x)11)
[Find ′ (x)]

= (x)ddx((ln 8x)11)+(ln 8x)11ddx(x)
[Use the Product Rule.]

= 11x(ln 8x)10ddx(ln 8x)+(ln 8x)11(1)
[Use the Chain Rule.]

= 11x(ln 8x)10(1x)+(ln 8x)11

= (ln 8x)10(11 + (ln 8x))

′ (x) = (ln 8x)10(11 + (ln 8x))

4.
If ($k$) = 7(ln $\sqrt{9{k}^{11}+31}\right)$, then find ′ ($k$).
 a. ($\frac{693}{2}$) ($\frac{{k}^{10}}{\sqrt{9{k}^{11}+31}}$) b. ($\frac{693}{2}$) ($\frac{{k}^{10}}{9{k}^{11}+31}$) c. ($\frac{{k}^{10}}{9{k}^{11}+31}$) d. 7 ln($\frac{99{k}^{10}}{2\sqrt{9{k}^{11}+31}}$) e. ()

#### Solution:

(k) = 7(ln 9k11+31)
[Write the function.]

′ (k) = ddk(7ln9k11+31)
[Find ′ (k).]

= 7(19k11+31)ddk(9k11+31)
[Use the Chain Rule.]

= 7(19k11+31)(129k11+31)ddk(9k11+31)
[Use the Chain Rule again.]

= 7(9)(11)k102(9k11+31) = (693 / 2) (k109k11+31)

′ (k) = (693 / 2) (k109k11+31)

5.
If $g$ ($x$) = ln ($x$ + $\sqrt{{x}^{2}+8}\right)$, then find $g$′ ($x$).
 a. ln(1 + $\frac{x}{\sqrt{{x}^{2}+8}}$) b. $\frac{1}{x}$ + $\frac{x}{\sqrt{{x}^{2}+8}}$ c. $\frac{1}{\sqrt{{x}^{2}+8}}$ d. $\frac{1}{x}$ - $\frac{x}{\sqrt{{x}^{2}+8}}$ e. $\frac{1}{x-\sqrt{{x}^{2}+8}}$

#### Solution:

g(x) = ln (x + x2+8)
[Write the function.]

g′(x) = ddx(ln(x+x2+8))
[Find g′(x).]

= 1x+x2+8ddx(x+x2+8)
[Use the Chain Rule.]

= 1x+x2+8(1 + (2x)2x2+8)
[Use the Chain Rule again.]

= 1x2+8

g′(x) = 1x2+8

6.
If $g$($t$) = , then find $g$′ ($t$).
 a. b. c. d. e. 22

#### Solution:

g(t) = (ln (2t+6))11
[Write the function.]

g′ (t) = ddt((ln (2t+6))11)
[Find g′ (t).]

= 11(ln (2t+6))10ddt(ln (2t+6))
[Use the General Power Rule and the Chain Rule.]

= 11(ln (2t+6))10(2(2t+6))
[Use the Chain Rule again.]

g′ (t) = 22(ln (2t+6))10(2t+6)

7.
If ($w$) = ln[cos (2$w$ + 7)], then find ′ ($w$).
 a. - tan (2$w$ + 7) b. - 2tan (2$w$ + 7) c. tan (2$w$ + 7) d. 2tan (2$w$ + 7) e. - $\frac{1}{\mathrm{cos}\left(2w+7\right)}$

#### Solution:

(w) = ln[cos (2w + 7)]
[Write the function.]

′ (w) = ddw(ln[cos (2w+7)])
[Find ′ (w).]

= 1cos (2w+7)ddw(cos (2w+7))
[Use the Chain Rule.]

= - 2(sin (2w+7)cos (2w+7))
[Use the Chain Rule again.]

= - 2tan (2w + 7)

′ (w) = - 2tan (2w + 7)

8.
If $g$($u$) = , then find (1440)$g$′ (0).
 a. b. 90(ln 10) - 96 c. 96 - 90(ln 10) d. 96 + 90(ln 10) e.

#### Solution:

g(u) = ln (8u+10)9u+12
[Write the function.]

g′ (u) = ddu(ln (8u+10)9u+12)
[Find g′ (u).]

= (9u+12)ddu(ln (8u+10))-(ln (8u+10))ddu(9u+12)(9u+12)2
[Use the Quotient Rule.]

= (9u+12)(88u+10)-(ln (8u+10))(9)(9u+12)2

= 8(9u+12)-(ln (8u+10))(9)(8u+10)(8u+10)(9u+12)2

g′ (u) = 8(9u+12)-9(8u+10)(ln (8u+10))(8u+10)(9u+12)2

g′ (0) = 96-90ln 101440
[Find g′ (0).]

(1440)g′ (0) = 96 - 90(ln 10)

9.
If ($l$) = (ln (9$l$ + 11))(ln (10$l$ + 12)), then find ′ (0).
 a. $\left(\frac{1}{12}\right)$ + $\left(\frac{1}{11}\right)$ b. $\frac{1}{132}$ c. $\frac{5}{6}$(ln 11) + $\frac{9}{11}$(ln 12) d. + e. 10(ln 9) + 9(ln 10)

#### Solution:

(l) = (ln (9l + 11))(ln (10l + 12))
[Write the function.]

′ (l) = ddl((ln (9l + 11))(ln (10l + 12)))
[Find ′ (l).]

= ln (9l + 11)ddl(ln (10l + 12)) + ln (10l + 12)ddl(ln (9l + 11))
[Use the Product Rule.]

= (ln (9l + 11))(1010l+12) + (ln (10l + 12))(99l+11)
[Use the Chain Rule.]

= 10(ln (9l + 11))10l+12 + 9(ln (10l + 12))9l+11

′ (0) = 5 / 6(ln 11) + 9 / 11(ln 12)
[Find ′ (0).]

10.
If ($k$) = ln(ln (6$k$ + 7)), then find ′ ($k$).
 a. b. c. d. $\frac{6}{\left(6k+7\right)}$ e.

#### Solution:

(k) = ln(ln (6k + 7))
[Write the function.]

′ (k) = ddk(ln(ln (6k+7)))
[Find ′ (k).]

= 1ln (6k+7)ddk(ln (6k+7))
[Use the Chain Rule.]

= 6(6k+7)(ln (6k+7))

′ (k) = 6(6k+7)(ln (6k+7))