INTRODUCING A USEFUL DODGE.
Sometimes one is stumped by finding that the expression to be differ- entiated is too complicated to tackle directly.
Thus, the equation
2 2 3
y = (x + a ) 2
is awkward to a beginner.
Now the dodge to turn the difficulty is this: Write some symbol, such as u, for the expression x2 + a2; then the equation becomes
3
y = u 2 ,
which you can easily manage; for
dy
=
du
3 1
u 2 .
2
Then tackle the expression
u = x2 + a2,
and differentiate it with respect to x,
du
= 2x.
dx
Then all that remains is plain sailing;
dy dy du
for
dx = du × dx ;
dy 3 1
that is,
dx = 2 u 2 × 2x
3 2 2 1
= 2 (x + a ) 2 × 2x
2 2 1
= 3x(x + a ) 2 ;
and so the trick is done.
By and bye, when you have learned how to deal with sines, and cosines, and exponentials, you will find this dodge of increasing useful- ness.
Examples.
Let us practise this dodge on a few examples.
Differentiate y = √a + x. Let a + x = u.
du 1
dy 1 − 1 1 − 1
= 1; y = u 2 ;
dx
du = 2 u 2 = 2 (a + x) 2 .
dy dy du 1
dx = du × dx = 2√a + x.
1
Differentiate y = √a + x2 .
Let a + x2 = u.
du − 1 dy
1 − 3
= 2x; y = u 2 ;
dx
du = −2 u 2 .
dy dy du x
dx = du × dx = −√(a + x2)3 .
Differentiate y =
2
m − nx3 +
p a
4 .
x3
2
Let m − nx3 + px
4
—
3 = u.
du 2 − 1
4 − 7
dx = −3 nx 3 − 3 px 3 ;
y = ua; dy
du
= aua−1.
=
×
dy dy du
2 p
a−1 1 7
dx du dx
4 3 3
= −a
( 2 nx− 3 + 4 px− 3 ).
x
3
m − nx3 +
1
Differentiate y = √x3 − a2 .
Let u = x3 − a2.
du 2
— 1 dy
1 3 2 − 3
= 3x ; y = u 2 ;
dx
du = −2 (x
— a ) 2 .
dy dy du
3×2
r
dx = du × dx = −2√(x3 − a2)3 .
Differentiate y = 1 − x.
1 + x
1
1
Write this as y = (1 − x) 2 .
(1 + x) 2
1 1
(1 + x) 1 d(1 − x) 2 − (1 − x) 1 d(1 + x) 2
dy 2
=
dx
dx
1 + x
1
2
dx .
— 1
(We may also write y = (1 − x) 2 (1 + x)
product.)
2 and differentiate as a
Proceeding as in example (1) above, we get
1
d(1 − x) 2 =
dx
1
—
2√1 − x
1
d(1 + x) 2
; and =
dx
1
2√1 + x.
Hence
1
dy (1 + x) 2
1
(1 − x) 2
dx = −2(1 + x)√1 − x − 2(1 + x)√1 + x
√
√ 1 √1 − x
= −2
dy
1 + x√1 − x − 2 (1 + x)3 ;
1
or dx = −(1 + x)√1 − x2 .
r x3
Differentiate y =
We may write this
1 + x2 .
3 2 − 1
y = x2 (1 + x ) 2 ;
3
1
2
— 1
3
2
dx = 2 x2 (1 + x )
2 + x2 ×
.
dx
dy d (1 + x2)− 1
—
Differentiating (1 + x2)
1
2 , as shown in example (2) above, we get
2
d (1 + x2)− 1 x
dx = −√(1 + x2)3 ;
so that
dy
3√x
√x5 √x(3 + x2)
dx = 2√1 + x2 − √(1 + x2)3 = 2√(1 + x2)3 .
Differentiate y = (x + √x2 + x + a)3. Let x + √x2 + x + a = u.
d (x2 + x + a)
1
du 2
= 1 + .
dx dx
y = u3; and dy
du
= 3u2 = 3 x + √x2 + x + a 2 .
1
Now let (x2 + x + a) 2 = v and (x2 + x + a) = w.
dw 1
dv 1 − 1
= 2x + 1; v = w 2 ;
dx
dw = 2 w 2 .
dv dv dw 1 2 − 1
dx = dw × dx = 2 (x + x + a) 2 (2x + 1).
du 2x + 1
Hence = 1 + √ , dx 2 x2 + x + a
dy dy du
dx = du × dx
r r
= 3 x + √x2 + x + a 2 1 +
2x + 1
—
2√x2 + x + a .
Differentiate y = We get
a2 + x2 3
a2 − x2
a2 x2 a2 + x2 .
2 2 1 2 2 1
1
1
) 6 (
x )
6 .
y = (a + x ) 2 (a − x ) 3 = (a2 + x2 1 a2 − 2 − 1
—
(a2 x2) 2 (a2 + x2) 3
2
2
1
6
dy d (a2 − x2)− 1
d (a2 + x2) 1
= (a
dx
+ x ) 6 +
dx
.
6
6
(a2 − x2) 1 dx
2
—
2
1
Let u = (a2 − x ) 6 and v = (a2 − x ).
— 1 du 1 − 7 dv
u = v 6 ; = −
v 6 ; = −2x.
dv 6 dx
du du dv 1 2
2 − 7
dx = dv × dx = 3 x(a − x ) 6 .
2 2 1 2 2
Let w = (a + x ) 6 and z = (a + x ).
1
w = z 6 ;
dw 1 5
—
=
z 6 ;
dz 6
dz
= 2x.
dx
dw dw dz 1 2
2 − 5
dx = dz × dx = 3 x(a
+ x ) 6 .
Hence
dy
2 2 1 x x
= (a + x ) 6 7 +
1 5 ;
dx 3(a2 − x2) 6 3(a2 − x2) 6 (a2 + x2) 6
dy x “s6 a2 + x2 1 #
or =
dx 3
(a2 − x2)7 + √6
.
(a2 − x2)(a2 + x2)5]
(9) Differentiate yn with respect to y5.
d(yn)
d(y5) =
nyn−1
5y5−1 =
nyn−5.
5
b
d (a − x)x }
√
(10) Find the first and second differential coefficients of y = x√(a − x)x.
dy x
=
dx b
1
2
+ (a − x)x.
dx b
2
Let (a − x)x 1 = u and let (a − x)x = w; then u = w 21 .
du 1 − 1 1 1
=
=
w
dw 2
dw
2
1
2w 2
= 2√(a − x)x.
dx = a − 2x.
du dw du a − 2x
dw × dx = dx = 2√(a − x)x.
Hence
dy x(a − 2x) √(a − x)x x(3a − 4x)
dx = 2b√(a − x)x + b = 2b√(a − x)x.
Now
d2y
2b√(a − x)x (3a − 8x) −
√
(3ax − 4×2)b(a − 2x)
(a − x)x
dx2 =
4b2(a − x)x
√
3a2 − 12ax + 8×2
= 4b(a − x) (a − x)x.
(We shall need these two last differential coefficients later on. See Ex. x. No. 11.)
Exercises VI. (See page 255 for Answers.) Differentiate the following:
(1) y = √x2 + 1. (2) y = √x2 + a2.
1 a
(3) y = √a + x. (4) y = √a − x2 .
(5) y =
√x2 a2
—
x2 . (6) y =
a2 + x2
√3 x4 + a
√2 x3 + a .
y = (a + x)2 .
Differentiate y5 with respect to y2.
√1 − θ2
Differentiate y =
.
1 − θ
The process can be extended to three or more differential coeffi-
dy dy dz dv
cients, so that = × × .
dx dz dv dx
Examples.
If z = 3×4; v = 7 ; y = √1 + v, find dv .
We have
dy
z2
1 dv
dx
14 dz 3
dv = 2√1 + v ;
dz = − z3 ;
= 12x .
dx
dy 168×3 28
dx = −(2√1 + v)z3 = −3×5√9×8 + 7 .
If t =
1
5√θ
; x = t3 +
t
; v = 2
7×2 dv
√3
, find .
x − 1 dθ
dv 7x(5x − 6) dx
2 1 dt 1
dx = 3√3
;
2
dv
(x − 1)4
= 3t
7x(5x − 6)(3t2 + 1 )
dt
+ 2 ;
dθ = −10√θ3 .
Hence
dθ = −
30√3
(x − 1)4√θ3 ,
an expression in which x must be replaced by its value, and t by its
value in terms of θ.
3a2x
√1 − θ2
√
1 dϕ
If θ =
We get
√x3 ; ω =
; and ϕ =
1 + θ
3 − ω√2 , find dx .
θ = 3a2x
1
—
2 ; ω =
1 − θ ; and ϕ = 1 + θ
1
√
3 − √2
ω−1.
r
dθ 3a2 dω 1
dx = −2√x3 ;
(see example 5, p. 68); and
dθ = −(1 + θ)√1 − θ2
dϕ 1
dθ 1
dω = √2ω2 .
1
3a2
So that = √ × √ × √ .
dx 2 × ω2 (1 + θ) 1 − θ2 2 x3
Replace now first ω, then θ by its value.
Exercises VII. You can now successfully try the following. (See page 256 for Answers.)
If u = 1 x3; v = 3(u + u2); and w = 1 , find dw .
2 v2 dx
If y = 3×2 + √2; z = √1 + y; and v = √
1
3 + 4z
dv
, find .
dx
x3
If y = √3
1 du
2 √
; z = (1 + y) ; and u = , find .
1 + z dx
