Download QUESTION (i) Express tanx as a quotient of trigonometric functions

QUESTION
(i) Express tan x as a quotient of trigonometric functions and hence, or
otherwise, derive the derivative of tan x with respect to x.
(ii) State the domain and range of the function x = tan−1 y and sketch its
graph.
(iii) Use the identity 1 + tan2 x = sec2 x to show that
d
1
.
(tan−1 y) =
dy
1 + y2
(iv) Obtain the first and second derivatives with respect to x of the function
u = x tan−1 (x3 ) and hence verify that u satisfies the equation
x2
du
d2 u
18x10
−
4x
.
+
4u
=
−
dx2
dx
(1 + x6 )2
ANSWER
(i) tan x =
sin x
cos x
d
cos x cos x − sin x(− sin x)
(tan x) =
dx
cos2 x
1
cos2 x + sin2 x
=
= sec2 x
=
cos2 x
cos2 x
(ii) If x = tan−1 y, then y = tan x.
y
y = tan x
x
π/2
x = tan−1 y
−π/2
0
π/2
0
x
y
−π/2
Domain −∞ < y < ∞
Range − π2 < x <
1
π
2
(iii) x = tan−1 y, y = tan x
dy
Therefore
= sec2 x = 1 + tan2 x = 1 + y 2
dx
d
1
1
dx
, or
=
(tan−1 y) =
Therefore
2
dy
1+y
dy
1 + y2
(iv) u = x tan−1 (x3 )
)
(
3x3
du
1
−1 3
−1 3
2
= tan (x ) + x
.3x = tan (x ) +
dx
1 + (x3 )2
1 + x6
d2 u
1
(1 + x6 )9x2 − 3x3 (6x5 )
2
=
.3x
+
dx2
1 + (x3 )2
(1 + x6 )2
3x2
9x2 (1 − x6 )
=
+
1 + x6
(1 + x6 )2
2
6
3x {1 + x + 3 − 3x6 }
=
(1 + x6 )2
3x2 (4 − 2x6 )
=
(1 + x6 )2
i.e.
d2 u
6x2 (2 − x6 )
=
dx2
(1 + x6 )2
Hence,
d2 u
du
−
4x
+ 4u
dx2
dx (
)
6x4 (2 − x6 )
3x3
−1 3
− 4x tan (x ) +
+ 4x tan−1 x3
(1 + x6 )2
1 + x6
6x4 (2 − x6 ) − 12x4 (1 + x6 )
(1 + x6 )2
4
6x (2 − x6 − 2 − 2x6 )
(1 + x6 )2
6x4 (−3x6 )
(1 + x6 )2
18x10
−
(1 + x6 )2
x2
=
=
=
=
=
2