Understanding Differentiation - JEE Main 2014 by ednexa.com

1. Differentiation
Differentiation of parametric functions
Sometimes x and y are given as functions of a
single variable, e.g., x = Φ (t), y = ψ (t) are two
functions and t is a variable. In such a case x
and y are called parametric functions or
parametric equations and t is called the
dy
parameter. To find dx in case of parametric
functions, we first obtain the relationship
between x and y by eliminating the parameter t
and then we differentiate it with respect to x.
But every time it is not convenient to eliminate
the parameter.
dy
Therefore
can also be obtained by the
dx
following formula

2. dy dy / dt

dx dx / dt
To prove it, let Δ x and Δ y be the changes in x
and y respectively corresponding to a small
change Δ t in t.
since
y y / t

,
x x / t
y dy
dy
y t0 t
 '(t)

 lim

 dt 
dx x 0 x lim x dx  '(t)
t 0 t
dt
lim

3. Multiple Choice Questions
1. If

t 

x  a  cos t  log  tan   , y  a sin t
2 


dy

dx
(a) tan t
(b) -tant
(c) cot t
(d) – cot t
then

4. Ans:
(a) tan t
t

x  a cos t  log tan 
Given that
2

and y  a sin t . Differentiating with respect
to t,
dy
 a cos t
we get dt
.....(i)

dx
 t  1 
2  t 
 a sin t  cot    sec  
and dt
 2  2 
 2 

1 
cos 2 t

 a  sin t 
 a cos t cot t
a
sin t 
sin t

.....(ii)
dy
From (ii) and (i), we get dx  tan t .

5. d2 y
2
2. If x  at , y  2at, then dx 2 
(a)

1
t2
1
 3
(c) t
1
 3
(b) 2at
1
(d)  2at 3

6. Ans:
1
 3
(d) 2at
dy dy / dt 2a dy 1 2a
 


y
dx dx / dt 2at , dx t
2
dy
d 2 y  dy 
y
 2a y 2     0
, dx  dx 
dx
d 2 y  (dy / dx )2
1
 2 

y
dx
2at 3 .
OR
d2y/dx2 = d/dx (dy/dx) =
[d/dt
(dy/dx)]
/
[dx/dt]
= [d/dt (1/t)] / [2at] = (-1/t2)/2at = 1/2at3

7. 3. If x  2cos t  cos 2t , y  2sin t  sin 2t ,
 dy
then at t  4 , dx 
(a)
2 1
(b)
(c)
2 1
2
(d) None of these
2 1

8. Ans:
(a)
2 1
dx
 2 sin t  2 sin 2t
dt
dy
 2 cos t  2 cos 2t
dt
dy cos t  cos 2t

dx
sin 2t  sin t

 dy 
t ,
 
Put
4 we have  dx  t  / 4
cos  / 4  cos  / 2

 2 1
.
sin  / 2  sin  / 4
and

9. Multiple Choice Questions
1. If
x  a sin 2 1  2cos   , y  bcos 2 1  cos 2  ,
dy

then dx
b tan 
(a) a
a tan 
(b) b
a
(c) b tan 
b
(d) a tan 
dy
3at
3at 2

x
,y 
, then
3
3
2. If
dx
1 t
1 t

t 2  t3
(a) 1  2t 3

t 2  t3
(c) 1  2t 3



t 2  t3
(b) 1  2t 3


t 2  t3
(d) 1  2t 3


10. 3. If x = a (t + sin t) and y = a (1 – cos t), then
dy

dx equals
(a) tan(t / 2)
(b) cot(t / 2)
(c) tan 2t
(d) tan t

11. Answers
1.
b tan 
(a) a
1


x  a sin 2  sin 4 
2

,
1


y  b cos 2  (1  cos 4 )
2



dx
 2a(cos 2  cos 4 )  2a.2 cos 3 cos 
d
and
dy
 2b(sin 4  sin 2 )  2b.2 cos 3 sin 
d

dy dy dx b


 tan 
.
dx d d a

12. 2.

t 2  t3
(b) 1  2t 3

3at
3at 2
x
, y
y  tx .
3
1t
1  t 3 . Clearly,
dy
dt
 t. 1  x .
Differentiate w.r.t. x, we get dx
dx
…..(i)
Now,
dx
1  t 3  t. 3 t 2 3a.(1  2 t 3 )
 3 a.

3 2
dt
(1  t )
(1  t 3 )2
.....(ii)
dy
3 at
(1  t 3 )2
t(2  t 3 )

t
.

3
3
dx
1  t 3 a(1  2 t ) 1  2 t 3 , {by (ii)}.

13. 3.
(a) tan(t / 2)
d
[a(1  cos t)]
dy dy / dt

 dt
d
dx dx / dt
[a(t  sin t)]
dt
t
t
2 sin cos
dy
a sin t
sin t
2
2



t
dx a  a cos t 1  cos t
2 cos 2
2

dy
t
 tan
dx
2.
Differentiation of infinite series
If y is given in the form of infinite series of x and
we have to find out
dy
dx
then we remove one or
more terms, it does not affect the series

14. i.
if y  f(x)  f(x)  f(x)  ....... ,
theny  f(x)  y  y  f(x)  y
2
2y
dy
dy
 f (x) 
dx
dx

,
dy
f (x)

,
dx 2y  1
ii.
f(x).....
If y  f(x)f(x)f(x)
 logy  y logf(x)
then y  f(x)y
1 dy y.f (x)
dy
dy
y 2f (x)

 logf(x). , 

y dx
f(x)
dx
dx f(x)[1 y logf(x)]
iii.
if y  f(x) 
1
f(x) 
1
f(x)  ....
then
dy
yf (x)

dx 2y  f(x)

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