318 Basic Engineering Mathematics
- y=
1
x
- y= 12
- y=x−
1
x^2
- y= 3 x^5 − 2 x^4 + 5 x^3 +x^2 − 1
- y=
2
x^3
- y= 4 x( 1 −x)
- y=
√
x 12. y=
√
t^3
- y= 6 +
1
x^3
- y= 3 x−
1
√
x
+
1
x
- y=(x+ 1 )^2 16. y=x+ 3
√
x
- y=( 1 −x)^2 18. y=
5
x^2
−
1
√
x^7
+ 2
- y= 3 (t− 2 )^2 20. y=
(x+ 2 )^2
x
- Find the gradient of the following curves at
the given points.
(a) y= 3 x^2 atx= 1
(b) y=
√
xatx= 9
(c) y=x^3 + 3 x−7atx= 0
(d) y=
1
√
x
atx= 4
(e) y=
1
x
atx= 2
(f) y=( 2 x+ 3 )(x− 1 )atx=− 2
- Differentiate f(x)= 6 x^2 − 3 x+5andfind
the gradient of the curve at
(a) x=−1(b)x= 2 - Find the differential coefficient of
y= 2 x^3 + 3 x^2 − 4 x−1 and determine the
gradient of the curve atx=2. - Determine the derivative of
y=− 2 x^3 + 4 x+7 and determine the
gradient of the curve atx=− 1. 5
34.6 Differentiation of sine and cosine functions
Figure 34.5(a) shows a graph ofy=sinx. The gradient
is continually changing as the curve moves from 0 to
y
0
(a)
(b)
0
y 5 sinx
x radians
x radians
1
2
2
A
A 9
09
C 9
B 9
D 9
BD
C
2
d
dx
1 dydx
2
2
3
2
3
2
(sinx) 5 cosx
2
Figure 34.5
AtoBtoCtoD. The gradient, given by
dy
dx
,maybe
plotted in a corresponding position belowy=sinx,as
shown in Figure 34.5(b).
At 0, the gradient is positiveand is at its steepest. Hence,
0 ′is a maximum positive value. Between 0 andAthe
gradient is positive but is decreasing in value until atA
the gradient is zero, shown asA′. BetweenAandBthe
gradient is negative but is increasing in value until atB
the gradient is at its steepest. HenceB′is a maximum
negative value.
If the gradient of y=sinx is further investigated
betweenBandCandCandDthen the resulting graph
of
dy
dx
is seen to be acosine wave.
Hence the rate of change of sinxis cosx,i.e.
ify=sinxthen
dy
dx
=cosx
It may also be shown that
ify=sinax,
dy
dx
=acosax (1)
(whereaisaconstant)
andify=sin(ax+α),
dy
dx
=acos(ax+α) (2)
(whereaandαare constants).
If a similar exercise is followed fory=cosxthen the
graphs of Figure 34.6 result, showing
dy
dx
to be a graph
of sinxbut displaced byπradians.