﻿ Derivatives of Trigonometric Functions Worksheet | Problems & Solutions

# Derivatives of Trigonometric Functions Worksheet

Derivatives of Trigonometric Functions Worksheet
• Page 1
1.
If $y$ = , then find $\frac{dy}{dc}$.
 a. (8 - 9cos 9$c$) b. (3$\sqrt{c}$) + c. d. (8$c$ - sin 9$c$)2((8 - 9cos 9$c$)(3$\sqrt{c}$) + ) e. 3

#### Solution:

y = (c(8c-sin 9c)3)
[Write the function.]

dydc = ddc(c(8c - sin 9c)3)
[Find dydc.]

= cDc((8c - sin 9c)3)+(8c - sin 9c)3)Dc(c)
[Use the Product Rule.]

= c(3(8c - sin 9c)2)Dc(8c - sin 9c)+(8c - sin 9c)312c
[Use the Chain Rule.]

= 3c(8c - sin 9c)2(8 - 9cos 9c)+(8c - sin 9c)32c

= (8c - sin 9c)2((8 - 9cos 9c)(3c) + 8c - sin 9c2c)

2.
If $y$ = 8$b$2sin ($\frac{9}{b}$), then find $\frac{dy}{db}$.
 a. (16$b$)sin($\frac{9}{b}\right)$ b. - 72cos$\left(\frac{9}{b}\right)$ c. - (16$b$)sin($\frac{9}{b}\right)$ + 72cos$\left(\frac{9}{b}\right)$ d. sin($\frac{9}{b}\right)$ - cos$\left(\frac{9}{b}\right)$ e. (16$b$)sin($\frac{9}{b}\right)$ - 72cos$\left(\frac{9}{b}\right)$

#### Solution:

y = 8b2sin(9b)
[Write the function.]

dydb = ddb(8b2sin(9b))
[Find dydb.]

dydb = (8b2)Db(sin(9b))+sin (9b)Db(8b2)
[Use the Product Rule.]

= (8b2)(cos(9b))Db (9b)+sin(9b)(16b)
[Use the Chain Rule.]

= - (8b2)cos(9b)(9b2) + (16b)sin(9b)

= (16b)sin(9b) - 72cos(9b)

dydb = (16b)sin(9b) - 72cos(9b)

3.
If $y$ = 12$\sqrt{\mathrm{cos}ec\sqrt{5x}}$, then find $\frac{dy}{dx}$.
 a. (3 ) ($\sqrt{\frac{5}{x}}\right)\left(\mathrm{cot}\sqrt{5x}\right)\left(\sqrt{\mathrm{cos}ec\sqrt{5x}}\right)$ b. ($\sqrt{\frac{5}{x}}\right)\left(\mathrm{cot}\sqrt{5x}\right)\left(\sqrt{\mathrm{cos}ec\sqrt{5x}}\right)$ c. - ($\sqrt{\frac{5}{x}}\right)\left(\mathrm{cot}\sqrt{5x}\right)\left(\sqrt{\mathrm{cos}ec\sqrt{5x}}\right)$ d. $\sqrt{\left(\mathrm{cot}\sqrt{5x}\right)\left(\mathrm{cos}ec\sqrt{5x}\right)}$ e. - (3 ) ($\sqrt{\frac{5}{x}}\right)\left(\mathrm{cot}\sqrt{5x}\right)\left(\sqrt{\mathrm{cos}ec\sqrt{5x}}\right)$

#### Solution:

y = 12cosec5x
[Write the function.]

dydx = ddx(12cosec5x)
[Find dydx.]

= 122cosec5xDx(cosec5x)
[Use the Chain Rule.]

= 122cosec5x(-(cosec5xcot5x)Dx(5x))
[Use the Chain Rule again.]

= - (6 ) cosec5x(cot5x)(52x)

= - (3 ) (5x)(cot5x)(cosec5x)

dydx = - (3 ) (5x)(cot5x)(cosec5x)

4.
If $y$ = , then find $\frac{dy}{dx}$.
 a. 5(5cos 5$x$+cos 10$x$) b. 25(7cos 5$x$+1) c. - 25(7cos 5$x$+1) d. 25(7cos 5$x$+cos 10$x$) e. - 25(7cos 5$x$+cos 10$x$)

#### Solution:

y = 5sin 5x7 + cos 5x
[Write the function.]

dydx = d dx( 5 sin 5 x7+cos 5 x)
[Find dydx.]

= 5((7+cos 5x)ddx(sin 5x)-sin 5xddx(7+cos 5x)(7+cos 5 x)2)
[Use the Quotient Rule.]

= 5((7 + cos 5x)(5cos 5x) - sin 5x(-5sin 5x)(7+cos 5x)2)

= 5 (35cos5x+5(1)(7+cos 5x)2)
[Use sin² θ + cos² θ = 1.]

= 25(7cos5x+1(7+cos 5x)2)
[Simplify.]

dydx= 25(7cos5x+1(7+cos 5x)2)

(7+cos 5x)2dydx= 25(7cos5x+1)

5.
If $g$($m$) = , then find $g$′ ($m$).
 a. (54 - 45sin $m$)(5cos $m$ + 6$m$)9 b. (54 + 45sin $m$)(5cos $m$ + 6$m$)8 c. 9(5cos $m$ + 6$m$)8 d. (6- 5 sin $m$)(5cos $m$ + 6$m$)8 e. (54 - 45sin $m$)(5cos $m$ + 6$m$)8

#### Solution:

g (m) = (5cos m + 6m)9
[Write the function.]

g′ (m) = ddm ((5cos m + 6m)9)
[Find g′ (m).]

g′ (m) = 9(5cos m+6m)8ddm(5cos m+6m)
[Use the Chain Rule.]

= 9(5cos m+6m)8(- 5sin m+6)

= (54 - 45sin m)(5cos m + 6m)8
[Simplify.]

g′ (m) = (54 - 45sin m)(5cos m + 6m)8

6.
If $g$($n$) = cos23$n$ sin46$n$, then find $g$′ ($n$) .
 a. 24(sin3 6$n$)(cos 6$n$)(cos23$n$) b. sin5 6n + (sin3 6n)(cos 6n)(cos23n) c. - 3(sin5 6$n$) + 24(sin3 6$n$)(cos 6$n$)(cos23$n$) d. 3(sin5 6$n$) e. 3(sin5 6$n$) - 24(sin3 6$n$)(cos 6$n$)(cos23$n$)

#### Solution:

g(n) = cos23n sin46n
[Write the function.]

g′ (n) = Dn(cos23n sin46n)
[Find g′ (n).]

= (sin46n)Dn(cos23n)+ (cos23n)Dn(sin46n)
[Use the Product Rule.]

= (sin46n)((2cos 3n)(-sin 3n)(3))+ 4sin 36n(cos 6n)(6)(cos23n)

= - 3(sin5 6n) + 24(sin3 6n)(cos 6n)(cos23n)
[Use 2sin θcos θ = sin 2θ and simplify.]

g′ (n) = - 3(sin5 6n) + 24(sin3 6n)(cos 6n)(cos23n)

7.
Find ′($u$), if ($u$)= 2$u$2 + 4cot ${u}^{3}$.
 a. 4$u$ + 12$u$2(cosec2 $u$3) b. 12$u$2(cosec2 $u$3) - 4$u$ c. 4$u$ - (cosec2 $u$3) d. 4$u$ - 12$u$2(cosec2 $u$3) e. 12$u$2(cosec2 $u$3)

#### Solution:

(u)= 2u2 + 4cot u3
[Write the function.]

′(u) = ddu(2u2 + 4cot u3)
[Find f ′ (u).]

= 4u - 4(cosec2 u3)(3u2)
[Use the Chain Rule.]

= 4u - 12u2(cosec2 u3)
[Simplify.]

′(u) = 4u - 12u2(cosec2 u3)

8.
If $g$($k$) = , then find ${D}_{k}$ [$g$($k$)].
 a. b. c. d. e. Does not exist

#### Solution:

g(k) = cosec 8k + 4
[Write the function.]

Dk (cosec 8k+4) = Dk((cosec 8k+4)12)
[Find Dk [g(k)] .]

= (-cosec 8kcot 8k)(8)2cosec 8k+4
[Use the Chain Rule.]

= -4(cosec 8k)(cot 8k)cosec 8k+4
[Simplify.]

Dk [g(k)] = -4(cosec 8k)(cot 8k)cosec 8k+4

9.
Find the derivative of the function
$y$ =

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

#### Solution:

y = 1sin2 2x+cos2 4x
[Write the function.]

dydx = ddx[1sin2 2x+cos2 4x]
[Find dydx.]

dydx = sin22x+cos24xDx(1)-Dxsin22x+cos24xsin22x+cos24x
[Use the Quotient Rule.]

= -12sin2 2x+cos2 4x 4sin 2x cos 2x-8cos 4x sin 4xsin2 2x+cos2 4x

= -12sin2 2x+cos2 4x(2sin 4x-4sin  8xsin2 2x+cos2 4x)
[Use 2sin θ cos θ = sin2 θ.]

= 2sin8x-sin4x(sin22x+cos24x)32

So, dydx = 2sin8x-sin4x(sin22x+cos24x)32

10.
If $f$($n$) = ${\mathrm{sin}}^{2}{\left(3{n}^{2}+2n\right)}^{3}$, then find $\frac{df}{dn}$.
 a. (18$n$ + 6)(cos 2(3$n$2 + 2n)3) b. (18$n$ + 6)(3$n$2 + 2$n$)2(sin 2(3$n$2 + 2$n$)3) c. sin (3$n$2 + 2$n$)3 d. (3$n$2 + 2$n$)2(sin 2(3$n$2 + 2$n$)3) e. (18$n$ + 6)(3$n$2 + 2$n$)2

#### Solution:

f(n) = sin2(3n2 + 2n )3
[Write the function.]

dfdn =ddn(sin2(3n2 + 2n )3)
[Find dfdn.]

= 2sin (3n2 + 2n)3ddn(sin (3n2 + 2n)3)
[Use the Chain Rule.]

= 2sin (3n2 + 2n)3cos (3n2 + 2n)3(3)(3n2 + 2n)2(6n + 2)
[Use the Chain Rule again.]

= (18n + 6)(3n2 + 2n)2(sin 2(3n2 + 2n)3)
[Simplify.]

dfdn = (18n + 6)(3n2 + 2n)2(sin 2(3n2 + 2n)3)