Higher Engineering Mathematics, Sixth Edition

(Nancy Kaufman) #1

Methods of differentiation 291


Thus


dy
dx

=( 5 )( 4 )x^4 −^1 +( 4 )( 1 )x^1 −^1 −

1
2

(− 2 )x−^2 −^1

+( 1 )

(

1
2

)
x−

1
2 −^1 − 0

= 20 x^3 + 4 +x−^3 −

1
2

x−

3
2

i.e.


dy
dx

= 20 x^3 + 4 +

1
x^3


1
2


x^3

Problem 6. Find the differential coefficients of
(a)y=3sin4x(b)f(t)=2cos3twith respect to
the variable.

(a) Wheny=3sin4xthen

dy
dx

=( 3 )(4cos4x)
=12cos4x

(b) When f(t)=2cos3tthen
f′(t)=( 2 )(−3sin3t)=−6sin3t


Problem 7. Determine the derivatives of
(a)y=3e^5 x(b)f(θ )=

2
e^3 θ

(c)y=6ln2x.

(a) Wheny=3e^5 xthen

dy
dx

=( 3 )( 5 )e^5 x=15e^5 x

(b) f(θ )=


2
e^3 θ

=2e−^3 θ, thus

f′(θ )=( 2 )(− 3 )e−^30 =−6e−^3 θ=

− 6
e^3 θ

(c) Wheny=6ln2xthen

dy
dx

= 6

(
1
x

)
=

6
x

Problem 8. Find the gradient of the curve
y= 3 x^4 − 2 x^2 + 5 x−2 at the points (0,−2)
and (1, 4).

The gradient of a curve at a given point is given by
the corresponding value of the derivative. Thus, since
y= 3 x^4 − 2 x^2 + 5 x− 2


Then the gradient=


dy
dx

= 12 x^3 − 4 x+ 5

At the point (0,−2),x= 0
Thus the gradient= 12 ( 0 )^3 − 4 ( 0 )+ 5 = 5


At the point (1, 4),x= 1
Thus the gradient= 12 ( 1 )^3 − 4 ( 1 )+ 5 = 13.


Problem 9. Determine the co-ordinates of the
point on the graphy= 3 x^2 − 7 x+2wherethe
gradient is−1.

The gradient of the curve is given by the derivative.

Wheny= 3 x^2 − 7 x+2then

dy
dx

= 6 x− 7

Since the gradient is−1then6x− 7 =−1, from which,
x= 1

Whenx=1,y= 3 ( 1 )^2 − 7 ( 1 )+ 2 =− 2
Hence the gradient is−1 at the point (1,−2).

Now try the following exercise

Exercise 115 Further problemson
differentiating common functions

In Problems 1 to 6 find the differential coeffi-
cients of the given functions with respect to the
variable.


  1. (a) 5x^5 (b) 2. 4 x^3.^5 (c)


1
x
[
(a) 25 x^4 (b) 8. 4 x^2.^5 (c)−

1
x^2

]


  1. (a)


− 4
x^2

(b) 6 (c) 2x

[
(a)

8
x^3

(b) 0 (c) 2

]


  1. (a) 2



x(b) 3

√ 3
x^5 (c)

4

x
[
(a)

1

x

(b) 5

√ 3
x^2 (c)−

2

x^3

]


  1. (a)


− 3

√ (^3) x (b)(x− 1 )
(^2) (c) 2sin3x



(a)
1
√ (^3) x 4 (b) 2(x− 1 )
(c) 6cos3x




  1. (a)−4cos2x(b) 2e^6 x(c)


3
e^5 x
[
(a) 8sin2x(b) 12e^6 x(c)
− 15
e^5 x

]
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