338 Higher Engineering Mathematics
Problem 6. Differentiatey=cot−^12 x
1 + 4 x^2Using the quotient rule:dy
dx=( 1 + 4 x^2 )(
− 2
1 +( 2 x)^2)
−(cot−^12 x)( 8 x)( 1 + 4 x^2 )^2
from Table 33.1(vi)=−2(1+ 4 xcot−^12 x)
(1+ 4 x^2 )^2Problem 7. Differentiatey=xcosec−^1 x.Using the product rule:dy
dx=(x)[
− 1
x√
x^2 − 1]
+(cosec−^1 x)( 1 )from Table 33.1(v)=− 1
√
x^2 − 1+cosec−^1 xProblem 8. Show that ify=tan−^1(
sint
cost− 1)
thendy
dt=1
2If f(t)=(
sint
cost− 1)then f′(t)=(cost− 1 )(cost)−(sint)(−sint)
(cost− 1 )^2=cos^2 t−cost+sin^2 t
(cost− 1 )^2=1 −cost
(cost− 1 )^2since sin^2 t+cos^2 t= 1=−(cost− 1 )
(cost− 1 )^2=− 1
cost− 1
Using Table 33.1(iii), wheny=tan−^1(
sint
cost− 1)thendy
dt=− 1
cost− 11 +(
sint
cost− 1) 2=− 1
cost− 1
(cost− 1 )^2 +(sint)^2
(cost− 1 )^2=(
− 1
cost− 1)(
(cost− 1 )^2
cos^2 t−2cost+ 1 +sin^2 t)=−(cost− 1 )
2 −2cost=1 −cost
2 ( 1 −cost)=1
2Now try the following exerciseExercise 135 Further problems on
differentiating inverse trigonometric
functions
In Problems 1 to 6, differentiatewith respect to the
variable.- (a) sin−^14 x (b) sin−^1
x
2
[
(a)4
√
1 − 16 x^2(b)1
√
4 −x^2]- (a) cos−^13 x(b)
2
3cos−^1x
3
[
(a)− 3
√
1 − 9 x^2(b)− 2
3√
9 −x^2]- (a) 3tan−^12 x(b)
1
2tan−^1√
x
[
(a)6
1 + 4 x^2(b)1
4√
x( 1 +x)]- (a) 2sec−^12 t(b) sec−^1
3
4x
[
(a)2
t√
4 t^2 − 1(b)4
x√
9 x^2 − 16]- (a)
5
2cosec−^1θ
2(b) cosec−^1 x^2
[
(a)− 5
θ√
θ^2 − 4(b)− 2
x√
x^4 − 1]- (a) 3 cot−^12 t(b) cot−^1
√
θ^2 − 1
[
(a)
− 6
1 + 4 t^2(b)
− 1
θ√
θ^2 − 1]