082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

10.5 Rate of change at a general point 367
At (1, 1), dy
dy
= 2 = gradient of tangent at A. At (−1,1),
dx dx
tangent atB. At (2, 4), dy = 4 = gradient of tangent atC.
dx
= −2 = gradient of
Suppose we wish to evaluate the derivative, dy
dx , at a specific value ofx, sayx 0
. This is
denoted by
dy
dx (x =x 0 )
dy
or more compactly by
dx (x 0 ) ory′ (x 0
)
An alternative notation is
dy
∣
dx
∣
x=x0
or
dy
∣
dx
∣
x0
So,for Example 10.7 wecould have written
dy
dx (1)=2
dy
dx (x=−1)=−2
dy
dx∣ =4 y ′ (2)=4
x=2
Example10.8 Refer to Figure 10.10. By considering the gradient of the tangent at the points A, B, C,
Dand Estatewhether dy ispositive, negative orzero atthese points.
dx
Solution AtAandCthetangentisparalleltothexaxisandso dy
dx iszero.AtBandEthetangent
has a positive gradient and so dy is positive. At D the tangent has a negative gradient
dx
and thus dy
dx isnegative.
y
E
B
A
D
C
x
Figure10.10
Graph forExample 10.8.
AswesawinChapter3,functionsareusedtorepresentphysicallyimportantquantities
suchasvoltageandcurrent.Whenthecurrentthroughcertaindeviceschanges,thiscan
give rise to voltages, the magnitudes of which are proportional to the rate of change of
the current. Consequently differentiation is needed to model these effects as illustrated
inEngineeringapplications10.1 and 10.2