MATHS
Method of differentiation
First principle of differentiation 1.
The derivative of a given function f at a point x = a in its domain represent the slope of the tangent at that point, and it is defined as:
Limit f (a h)f (a ) , provided the limit exists & is denoted by f (a). h0 h i.e. f (a) = Limit x a 2.
f ( x )f (a) , provided the limit exists. xa
If x and x + h belong to the domain of a function f defined by y = f(x), then
Limit f ( x h)f ( x ) if it exists, is called the Derivative of f at x & is denoted by f (x) or dy . h0 dx h f ( x h)f ( x ) h This method of differentiation is also called ab-initio method or first principle method. i.e., f (x) = Limit h0
Example # 1 : Find derivative of following functions by first principle with respect to x. (i) f(x) = x 2 (ii) f(x) = tan x (iii) f(x) = esinx Solution :
(i)
(ii)
f(x)
( x h) 2 x 2 = hlim 0 h
f(x)
sin( x h) sin x tan( x h) tan x lim lim cos( x h) cos x = h0 = h0 h h
= hlim 0
(iii)
f(x)
2xh h2 = hlim = 2x. 0 h
sin( x h – x ) = sec 2x. h cos x. cos( x h)
e sin ( x h) e sin x = hlim 0 h
e sin ( x h ) sin x 1 sin( x h) sin x lim sin x = h0 e h sin( x h) sin x
lim sin( x h) sin x = esin x lim h0 h
= esin x
h0
= esin x
h0
lim
xh x xhx 2 cos sin 2 2 h
h sin(h / 2) sin x cos x cos x 2 h/2 = e
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MATHS Derivative of some elementary functions : f(x)
f(x)
1.
xn
nx n – 1 ,
(x R, n R)
2.
ax
ax n a ,
a>0
3.
n |x|
1 x
4.
logax
1 x n a
5.
sin x
cos x
6.
cos x
– sin x
7.
sec x
sec x tan x
8.
cosec x
– cosec x cot x
9.
tan x
sec 2 x
10.
cot x
– cosec 2x
Basic theorems : Sum of two differentiable functions is always differentiable. Sum of two non-differentiable functions may be differentiable. There are certain basic theorems in differentiation: 1.
d (f ± g) = f(x) ± g(x) dx
2.
d d (k f(x)) = k f(x) dx dx
3.
d (f(x) . g(x)) = f(x) g(x) + g(x) f(x) dx
4.
d dx
5.
d (f(g(x))) = f(g(x)) g(x) dx
g( x ) f ( x ) f ( x ) g( x ) f ( x) = g2 ( x ) g( x )
This rule is also called the chain rule of differentiation and can be written as dy dy dz = . dx dz dx
Note that an important inference obtained from the chain rule is that
dy dy dx =1= . dx dy dy
1 dy = dx / dy dx
another way of expressing the same concept is by considering y = f(x) and x = g(y) as inverse functions of each other.
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MATHS dy = f(x) dx
g(y) =
dx dy = g(y)
and
1 f ( x )
Example # 2 : Find the derivative of the following functions with respect to x.
Solution :
sin( 2 x 3)
(i)
f(x) =
(ii)
f(x) =
(i)
f(x) =
sin ( 2 x 3 )
f(x) =
d 1 d ( sin ( 2 x 3 ) ) = . (sin (2x + 3)) dx 2 sin (2x 3) dx
x
(iii)
1 x2
f(x) = x . sin x
(chain rule)
cos(2x 3) =
(ii)
f(x) =
sin (2x 3) x 1 x2
= (iii)
f(x)
=
(1 x 2 ) x(2x ) (1 x 2 )2
(Quotiant rule)
1 x2 (1 x 2 )2
f(x) = x sin x f(x) = x. cos x + sin x
(Product rule)
Example # 3 : If f(x) = sin (x + tanx), then find value of f(0). Solution :
f(x) = sin (x + tanx) f(x) = cos (x + tanx) (1 + sec 2x) Hence, f(0) = 2
(chain rule)
Self Practice Problems : (1)
Find the derivative of following functions using first principle with respect to x. (i) f(x) = x sin x (ii) f(x) = sin2 x
(2)
f (5 t ) f (5 t ) If f(5) = 7, then find the value of tlim 0 2t
(3)
Differentiate the following functions with respect to x. (i)
(iii)
(1 + 3x 2) (2x 3 – 1)
1 x 2
(ii)
(iv)
( x 1) ( x 2)( x 3) 1 x 1 x
(v)
cos 3 x sin x
(vi)
x ex sin x
(vii)
sin x 1 cos x
(viii)
n (sin x – cos x)
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MATHS Answers :
(1)
(i)
(2)
7
(3)
(i)
x cosx + sinx
(ii)
2sin x cos x
6x (5x 3 + x – 1)
(ii)
( x 2) 2 ( x 3 ) 2
(iv)
(1 x )1 / 2 (1 x )3 / 2
x 2 2x 1
x (iii)
1 x
1 2
(v)
cos 4 x – 3 cos 2x sin2x (vi)
ex ((sin x + cos x) x + sin x)
(vii)
1 x sec 2 2 2
cos x sin x sin x cos x
(viii)
Derivative of inverse trigonometric functions : 1.
–
dx = cos y dy
dy 1 = = dx cos y
dy = dx
1
, siny ± 1
1 sin 2 y
1 1 x 2
, – 1 < x < 1.
In general |cos y| =
2.
y and x = sin y 2 2
y = sin–1 x
y = tan–1x
1 sin 2 y . But here since –
x = tan y and –
y 2 2
|cos y| = cosy
y 2 2
dx = sec 2y = 1 + tan2 y dy
3.
dx 2 dy = 1 + x
y = sec –1x
dx dy = sec y tan y
dy 1 = dx 1 x2
(x R)
y [0, ] – and x = secy 2
1 dy = sec y. tan y , tan y 0 dx
1 = | sec y. tan y | , y 0, , 2 2 1 1 = | sec y || tan y | = | x | x2 1
1 dy = dx | x | x2 1
x (– , – 1) (1, )
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MATHS Results for the derivative of inverse trigonometric functions can be summarized as : f(x) f(x) 1 sin–1x ; |x| < 1 1 x 2 1 cos –1x ; |x| < 1 1 x 2 1 tan–1x ; xR 1 x2 1
cot–1x
xR
;
1 x2
1
sec –1 x
;
| x | x2 1
|x| > 1
1 cosec -1 x
; |x| > 1
| x | x2 1
Example # 4 : If f(x) = n (sin–1 x 2), then find f(x).
Solution :
f(x)
=
1 (sin 1 x 2 )
2x
1
.
. 2x =
1 ( x 2 )2
(sin 1 x 2 ) 1 x 4
Example # 5 : If f(x) = 2x sec –1x – cosec –1(x) , then find f(–2). Solution :
f(x) = 2 sec –1(x) +
2x | x | x2 1
Hence, f(–2) = 2.sec –1(– 2) –
f(–2) =
+
1 | x | x2 1
2 3
1 +
2 3
4 3 – 3 2
Logarithmic differentiation : The process of taking logarithm of the function first and then differentiate is called Logarithmic differentiation. It is often useful in situations when (i) (ii)
a function is the product or quotient of a number of functions OR a function is of the form [f(x)]g(x) where f & g are both derivable,
Example # 6 : If y = (sin x)n x, find Solution :
dy dx
n y = n x . n (sin x)
1 dy cos x 1 y dx = x n (sin x) + n x. sin x
n sin x dy cot x n x = (sin x)n x x dx
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MATHS Example # 7 : If y =
Solution :
x1/ 2 (1 2x )2 / 3 ( 2 3 x )3 / 4 (3 4 x ) 4 / 5 y=
, then find
dy . dx
x1/ 2 (1 2x )2 / 3 ( 2 3 x )3 / 4 (3 4 x ) 4 / 5
taking loge on both side n y =
2 4 1 3 n x + n (1 – 2x) – n (2 – 3x) – n (3 – 4x) 3 5 2 4
1 dy 9 16 4 1 = – + + y dx 4 (2 3 x ) 5 (3 4 x ) 3(1 2x ) 2x
1 4 9 16 dy = y 2x 3 (1 2 x ) 4(2 3 x ) 5 (3 4 x ) dx
Implicit differentiation : If f(x, y) = 0, is an implicit function i.e. y can't be expressed explicitly as a function of x then in order to find dy/dx, we differentiate each term w.r.t. x regarding y as a function of x and then collect terms in dy/dx. Example # 8 : If x 3 + y3 = 3xy, then find Solution :
dy . dx
Differentiating both sides w.r.t.x, we get 3x 2 + 3y2
dy dy = 3x + 3y dx dx
y x2 dy = 2 y x dx Note that above result holds only for points where y2 – x 0 Example # 9 : If x y + yx = 2 then find Solution :
dy dx
u+v=2 du dv + =0 dx dx where u = x y n u = y n x
........(i) & &
v = yx n v = x n y
1 du dy y = + n x & u dx dx x
x dy 1 dv = n y + y dx v dx
dy y du & = x y n x dx dx x
x dy dv = yx n y y dx dx
Now, equation (i) becomes
x dy dy y + y x n y x y n x y dx = 0. dx x
x y y y n y x . dy x =– dx y x x n x y x . y
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MATHS Self Practice Problems (4)
(5)
Differentiate the following functions with respect to x. (i)
y = sec –1 (x 2)
(iii)
1 y = 1 x
(v)
y = (ln x)x + (x)sin x
Find (i) (iii)
(ii)
1 x y = tan–1 1 x
(iv)
y = ex
(ii)
x 2/3 + y2/3 = a2/3
x
dy if dx y = cos (x + y) x = y n (x – y)
(6)
If x y = ex – y, then prove that
(7)
If
x
n x dy = . dx (1 n x )2
x a x dy = log , then prove that =2– . xy xy y dx
Answers :
(4)
(i)
2
(ii)
x x 1 4
1 (iii) 1 x
x
1 1 n 1 x 1 x
1 1 x2 x
(iv) x x. e x (nx + 1)
1 sin x cos x nx (v) n (nx ) nx (n x)x + x sinx x sin( x y ) (i) 1 sin( x y )
(5)
y (ii) – x
1/ 3
(iii)
y( x 2 y ) x( x y ) y 2
Differentiation using substitution : In certain situations as mentioned below, substitution simplifies differentiation. For each of the following expression, appropriate substitution is as follows (i)
x 2 a2
x = a tan , where
<< 2 2
or
x = a cot , where 0 < <
(ii)
a2 x 2
x = a sin , where
2 2
or
x = a cos , where 0
(iii)
x 2 a2
x = a sec , where [0 – 2
or
(iv)
xa ax
x = a cosec , where , – {0} 2 2
x = a cos , where 0
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MATHS 1 x 2 1 Example # 10 : Differentiate y = tan–1 with respect to x. x Solution :
Let
x = tan , where 2 , 2 – {0} | sec | sec , 2 2
| sec | 1 y = tan–1 tan
1 cos y = tan–1 sin
y = tan–1 tan 2
y=
2
y=
1 tan–1 x 2
Example # 11 : Find Solution :
1 tan (tan x ) x for x , 2 2 1 dy = 2(1 x 2 ) dx
dy , where y = tan–1 dx
1 x 1 x 1 x 1 x
Let
x = cos , where [0, ]
1 cos 1 cos y = tan–1 1 cos 1 cos
y = tan–1
2 cos 2 sin 2 2 2 cos 2 sin 2 2
y = tan–1
y=
.
1 cos
1 tan 1 tan
2 cos
2 2
1 – cos –1x 4 2
but for 0, , 2 2 2
– 4 2
y=
1 dy = dx 2 1 x2
2 cos
as
2 cos 2 2
– 4 4 2 4
2x Example # 12 : If f(x) = sin–1 , then find 1 x2
(i)
f(2)
(ii)
1 f 2
(iii)
f(1)
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MATHS Solution :
x = tan ,
where –
<< 2 2
y = sin–1 (sin 2)
, 2 2 2 , 2 y = 2 2 2 ( 2) , 2 2 2 2 1 x 2 2 f(x) = 1 x 2 1 x 2
2 tan 1 x x 1 1 2 tan x 1 x 1 f(x) = ( 2 tan 1 x ) x 1
x 1 1 x 1 x 1
2 5
(ii)
1 8 f = 2 5
(i)
f(2) = –
(iii)
f(1+) = – 1 and f(1–) = + 1 f(1) does not exist.
Aliter Above problem can also be solved without any substution also, but in a little tedious way. 2x f(x) = sin–1 1 x2
1
f(x) = 1
.
4x 2
(1 x 2 )2
thus
2(1 x 2 ) (1 x 2 )2
(1 x ) 2
.
| 1 x 2 |
2 , |x|1 2 1 x f(x) = 2 , | x|1 1 x 2
Example # 13 : If
Solution :
.
(1 x 2 )
2 f(x) =
(1 x 2 )2
(1 x 2 )2
(1 x 2 ) =
2{(1 x 2 ) 2x 2 }
Put
1 x 2 +
1 y 2 = a(x – y), then prove that
x = sin where –
2 2
dy = dx
1 y 2 1 x 2
.
and
. 2 2 cos + cos = a (sin – sin) y = sin where –
cos = 2a cos sin 2cos 2 2 2 2
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MATHS =a cot 2
– = 2 cot–1 (a) sin–1 x – sin–1 y = 2 cot–1a differentiating w.r.t to x.
1 1 x
2
1 y
2
dy =0 dx
1 y 2
dy = dx
1
–
– + 2 cot–1a – + 2 cot–1a
or or
1 x 2
Aliter Using implicit differentiation
x 1 x
dy = dx
y 2
–
1 y
dy dy = a 1 dx dx
2
y a 1 y2
dy =a+ dx
x 1 x 2
dy = dx
a
x 1 x2 y 1 y2
1 x 2 1 y 2 x xy 1 x 2 1 x2 1 y2 y xy 1 y2
(1 x 2 ) (1 x 2 )(1 y 2 ) x 2 xy dy = . dx (1 x 2 )(1 y 2 ) (1 y 2 ) xy y 2
dy = dx
a
1 y 2 1 x 2
1 (1 x 2 )(1 y 2 ) xy
1 y2 1 x2
=
1 (1 x 2 )(1 y 2 ) xy
1 y2
.
1 x2
Hence proved
Parametric differentiation : dy dy / d If y = f() & x = g() where is a parameter, then dx dx / d . Example # 14 : If x= a cos 3t and y = a sin3t, then find the value of
Solution :
dy . dx
3a sin 2 t cos t dy dy / dt = = = – tan t 3a cos 2 t sin t dx dx / dt
Example # 15 : If y = a cos t and x = a (t – sint), then find the value of
Solution :
dy a sin t = dx a(1 cos t )
dy dx
t
2
dy at t = . dx 2
= – 1.
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MATHS Derivative of one function with respect to another function : Let y = f(x); z = g(x) then
dy dy / dx f '(x) . dz dz / dx g' (x)
Example # 16 : Find derivative of y = n x with respect to z = ex. Solution :
1 dy dy / dx = = dz dz / dx xex
Self Practice Problems :
(8)
Find
dy when dx
(i)
x = a (cos t + t sin t)
and
y = a (sin t – t cos t)
(ii)
1 t2 x=a 2 1 t
and
2t y=b 1 t 2
sin–1
x2 4 4 x a
2 xa 2 dy , then prove that = . dx x 4 a4
(9)
If y =
(10)
2x dy 2 If y = tan–1 = (| x | 1) 2 , then prove that dx 1 x 1 x2
(11)
If u = sin (m cos –1x) and v = cos (m sin–1 x), then prove that
Answer :
(8)
(i)
tan t
du = dv
1 u2 1 v 2
.
( t 2 1)b 2at
(ii)
Higher order derivatives : Let a function y = f(x) be defined on an open interval (a, b). It’s derivative, if it exists in (a, b) is a certain function f (x) [or (dy/dx) or y ] & is called the first derivative of y w. r. t. x. If it happens that the first derivative has a derivative in (a, b) then this derivative is called the second derivative of y w. r. t. x & is denoted by f (x) or (d2y/dx 2) or y . While the first derivative denotes slope of the graph, the second derivative denotes it's concavity. Similarly, the 3rd order derivative of y w. r. t. x, if it exists, is defined by
d3 y dx 3
d d2 y dx dx 2 , it is also
denoted by f (x) or y . It must be carefully noted that in case of parametric functions although
dy dy / dt = dx dx / dt
but
d2 y dx
2
d2 y / dt 2 2
dx / dt
2
rather
d2 y dx
2
=
d dy / dt dx dx / dt
which on applying chain rule can be resolved as
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MATHS
d2 y dx 2
d2 y dx 2
d dy / dt dt . = dt dx / dt dx
=
d2 y
dx 2
=
dx d2 y dy d2 x . . dt dt 2 dt dt 2 dx dt
2
.
dt dx
dx d2 y dy d2 x . . 2 dt dt 2 dt dt dx dt
3
Example # 17 : If y = x 3 n x, then find yand y Solution :
y = 3x 2 n x + x 3
1 = 3x 2 n x + x 2 x
y = 6x n x + 3x 2 .
1 + 2x = 6x n x + 5x x
y = 6 n x + 11
x
1 Example # 18 : If y = , then find y(1) x Solution :
Now taking loge both sides, we get n y = – x n x when x = 1, then y = 1 n y = – x n x
y y = – (1 + n x)
y = – y (1 + n x)
y = y (1 + n x)2 –
......(i)
again diff. w.r.t. to x, y = – y(1 + n x) – y .
1 x
y(1) = 0
Example # 19 : If x = t + 1 and y = t2 + t3, then find
Solution :
dy = 2t + 3t2 ; dt
y (using (i)) x
d2 y dx 2
.
dx =1 dt
dy = 2t + 3t2 dx
d2 y dx
2
d2 y dx 2
=
d dt (2t + 3t2) . dt dx
= 2 + 6t.
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MATHS Example # 20 : Find second order derivative of y= sin x with respect to z = ex. Solution :
cos x dy dy / dx = = dz dz / dx ex
d2 y
dz
= d2 y dz
2
2
=
d cos x dz e x
e x sin x cos x e x x 2
(e )
=–
.
d2 y dx
2
=
d cos x dx . dx e x dz
1 ex
(sin x cos x ) e2x
Example # 21 : y = f(x) and x =g(y) are inverse functions of each other, then express g(y) and g(y) in terms of derivative of f(x). dx dy Solution : = f(x) and dy = g(y) dx 1 g(y) = ...........(i) f ( x ) again differentiating w.r.t. to y
d g(y) = dy
=
d dx
=–
1 f ( x )
1 dx . f ( x ) dy
f ( x ) f ( x )2
. g(y)
f ( x ) g(y) = –
f ( x )3
.........(ii)
which can also be remembered as
d2 y d x dx 2 2 = – 3 . dy dy dx 2
Example # 22 : y = sin (sinx) then prove that y + (tanx) y + y cos 2x = 0 Solution :
Such expression can be easily proved using implicit differentiation y = cos (sin x) cos x sec x.y = cos (sin x) again differentiating w.r.t x, we get secx y + y sec x tan x = – sin (sin x) cos x y + y tan x = – y cos 2 x y +(tanx) y + y cos 2x = 0
Self Practice Problems :
(12)
d2 y n x If y = , then find the value of . x dx 2
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MATHS cos 2
(13)
Prove that y = x + tan x satisfies the differential equation
(14)
If x = a (cos + sin ) and y = a(sin – cos), then find the value of
(15)
Find the second order derivative of nx with respect to sin x.
(16)
If y =
e– x
Answers :
(A cos x + B sin x), then prove that
(12)
2n x 3 x
(14)
3
d2 y dx 2
+2.
sec 3 a
x
d2 y dx 2
– 2y + 2x = 0.
d2 y dx 2
.
dy + 2y = 0. dx
(15)
x sin x cos x x 2 cos3 x
Derivative of a determinant :
If F(x) =
f (x)
g( x )
h( x )
l( x )
m( x )
n( x )
u( x )
v( x )
w( x)
, where f, g, h, l, m, n, u, v, w are differentiable functions of x, then
f ' ( x ) g' ( x ) h' ( x ) F (x) =
l( x )
m( x )
n( x )
u( x )
v( x )
w( x)
f (x) +
g( x )
h( x )
l' ( x ) m' ( x ) n' ( x ) u( x )
v( x )
w( x )
+
f ( x)
g( x )
h( x )
l( x )
m( x )
n( x )
u' ( x ) v' ( x ) w ' ( x )
L’ hospital’s rule : If f(x) & g(x) are functions of x such that: (i)
Limit f(x) = x a
= Limit g(x) x a
OR (ii)
Limit f(x) = 0 = Limit g(x), both f(x) and g(x) are continuous at x = a, both f(x) and x a
x a
g(x) are differentiable at x = a and both f (x) and g (x) are continuous at x = a, then
Limit x a
f (x) f ' (x) Limit g( x ) = x a g' ( x )
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