Higher Engineering Mathematics, Sixth Edition

(Nancy Kaufman) #1

292 Higher Engineering Mathematics



  1. (a) 4ln9x (b)


ex−e−x
2

(c)

1 −

√ x x ⎡ ⎢ ⎢ ⎣

(a)

4
x

(b)

ex+e−x
2

(c)

− 1
x^2

+

1
2


x^3






  1. Find the gradient of the curve y= 2 t^4 +
    3 t^3 −t+4 at the points (0, 4) and (1, 8).
    [−1, 16]

  2. Find the co-ordinates of the point on the
    graphy= 5 x^2 − 3 x+1 where the gradient
    is 2.


[( 1
2 ,

3
4

)]


  1. (a) Differentiatey=
    2
    θ^2


+2ln2θ−

2 (cos 5θ+3sin2θ)−

2
e^3 θ

(b) Evaluate

dy

in part (a) whenθ=

π
2

,
correct to 4 significant figures.





(a)

− 4
θ^3

+

2
θ

+10sin5θ

−12cos2θ+

6
e^3 θ
(b) 22. 30







  1. Evaluate


ds
dt

, correct to 3 significant figures,

whent=

π
6

given
s=3sint− 3 +


t. [3.29]

27.5 Differentiation of a product

Wheny=uv,anduandvare both functions ofx,

then

dy
dx

=u

dv
dx

+v

du
dx
This is known as theproduct rule.

Problem 10. Find the differential coefficient of
y= 3 x^2 sin2x.

3 x^2 sin2xis a product of two terms 3x^2 and sin2x
Letu= 3 x^2 andv=sin2x

Using the product rule:
dy
dx

= u

dv
dx

+ v

du
dx
↓↓ ↓ ↓
gives:

dy
dx

=( 3 x^2 )(2cos2x)+(sin 2x)( 6 x)

i.e.

dy
dx

= 6 x^2 cos2x+ 6 xsin2x

= 6 x(xcos 2x+sin 2x)

Note that the differential coefficient of a product is
notobtained by merely differentiating each term and
multiplying the two answers together. The product rule
formulamustbe used when differentiating products.

Problem 11. Find the rate of change ofywith
respect toxgiveny= 3


xln2x.

The rate of change ofywith respect toxis given by

dy
dx

y= 3


xln2x= 3 x

1

(^2) ln2x, which is a product.
Letu= 3 x
1
(^2) andv=ln2x
Then
dy
dx
= u
dv
dx




  • v
    du
    dx
    ↓↓ ↓ ↓


    (
    3 x
    1
    2
    )(
    1
    x
    )
    +(ln2x)
    [
    3
    (
    1
    2
    )
    x
    1
    2 −^1
    ]
    = 3 x
    1
    2 −^1 +(ln2x)
    (
    3
    2
    )
    x−
    1
    2
    = 3 x−
    1
    2
    (
    1 +
    1
    2
    ln2x
    )
    i.e.
    dy
    dx


    3

    x
    (
    1 +
    1
    2
    ln 2x
    )
    Problem 12. Differentiatey=x^3 cos3xlnx.
    Letu=x^3 cos3x(i.e. a product) andv=lnx
    Then
    dy
    dx
    =u
    dv
    dx
    +v
    du
    dx
    where
    du
    dx
    =(x^3 )(−3sin3x)+(cos 3x)( 3 x^2 )
    and
    dv
    dx


    1
    x



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