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

374 Chapter 10 Differentiation
(f) Note that 1 x 5 =x−5 . UsingTable 10.1, we find thatify =x −5 theny ′ = −5x −6 .
(g) FromTable 10.1, ify = cosh −1 (ax +b) then
y ′ =
a
√
(ax+b)2 −1
Inthiscase,a = 5 andb= 0.Hence, if
y = cosh −1 5x then y ′ =
5
√
25x2 −1
Example10.12 Differentiatey(t) = e t .
Solution We note that the independent variable ist. However, Table 10.1 can still be used. From
Table 10.1, wefind
dy
dt =et =y
We note that the derivative of e t is again e t . This is the only function which reproduces
itself upon differentiation.
EXERCISES10.7
1 UseTable 10.1 to findy ′ given:
(a) y=t 2 (b) y=t 9
(c) y=t −3 (d) y=t
(e) y= 1 (f) y= 1
t t 2
(g) y = e 3t (h) y = e −3t
(i) y= 1
e 5t (j) y =t 1/2
(k) y =sin(2t +3) (l) y =cos(4 −t)
( )
t
(m) y = tan
2 +1 (n) y = cosec(3t +7)
(o) y =cot(1−t)
(q) y =sin −1 (t +π)
(p) y =sec(2t −π)
(r) y = π
(s) y = tan −1 (−2t −1) (t) y = cos −1 (4t −3)
(u) y = tanh(6t)
( )
(v) y = cosh(2t +5)
t +3
(w) y = sinh
2
(x) y = sech(−t)
( )
2t
(y) y = coth
3 − 1 2
(z) y = cosh −1 (t +3)
2 Find dy
dx when
(a) y= 1 √ x
(b) y = e 2x/3
(c) y = e −x/2 ( )
2x−1
(e) y = cosec
3
(f) y = tan −1 (πx +3)
(g) y = tanh(2x +1)
(h) y = sinh −1 (−3x)
(i) y = cot(ωx + π)
1
(j) y=
sin(5x +3)
(l) y= 1
cos3x(
)
x −1
(n) y = cosech
2
(o) y = tanh −1 (
2x+3
7
)
(d) y = lnx
ω constant
(k) y = cos3x
(m) y = tan(2x + π)