SHORT NOTES MODULE 1 : THE DERIVATIVES
LIMIT EXISTENCE OF LIMIT
LIMIT OF f(x) AT x=a EXISTS IF AND ONLY IF
LEFT HAND LIMIT EQUALS RIGHT HAND LIMIT
lim f ( x ) = lim f ( x ) = L
lim f ( x ) = L
x→a+
x→a
x→a−
EVALUATION OF LIMIT
Limit at a number:
f(x) is a polynomial
f(x) is a rational function
f(x) is a rational function
f(x) is a rational function
SUBSTITUTION
SUBSTITUTION
CONJUGATION
FACTORIZATION
f(x) is a rational function LIMIT DOES NOT EXIST
LIMIT AT INFINITY
lim f ( x )
f(x) is a polynomial LEADING TERM
x →
lim an xn + an−1xn−1 +
a0 = lim an xn
x →
f(x) is a rational function; đ?‘“ đ?‘Ľ =
x →
đ?‘?(đ?‘Ľ) đ?‘ž(đ?‘Ľ)
LEADING TERM OF p(x) AND q(x)
an xn + an−1xn−1 +
lim
x → b xm m
+ a0
= lim
an xn
an xn + an−1xn−1 +
+ a0
+ bm−1xm−1 +
+ b0
x → b xm m
đ?‘?(đ?‘Ľ)
f(x) is a rational function; đ?‘“ đ?‘Ľ = đ?‘ž(đ?‘Ľ)
DIVIDE p(x) AND q(x) BY HIGHEST TERM OF q(x) an x + an−1x n
lim
x → b xm m
n−1
+
+ bm−1xm−1 +
+ a0
xm = lim m + b0 x→ bm x + bm−1xm−1 +
+ b0
xm đ?‘˜
f(x) is a rational function; đ?‘“ đ?‘Ľ = đ?‘ž(đ?‘Ľ) ZERO lim
x → b xm m
k + bm−1x
m−1
+
+ b0
CONTINUITY
f(x) is continuous at x = a if and only if the following conditions hold:
f ( a ) = lim f ( x )
lim f ( a ) = k
f (a) = k
x →a
x →a
&
=k
&
If at least one condition fails, then f(x) is not continuous at x = a.
DIFFERENTIATION FIRST PRINCIPLE OF DERIVATIVE FIRST PRINCIPLE
= lim
k
x → b xm m
=0
RULES OF DIFFERENTIATION y=a
EQUATION
RULE/TYPE OF FUNCTION Constant
y = xn
Power
y = af ( x ) bg ( x )
Addition/Subtraction
y = uv
y=
Product
u v
y = f ( x )
Quotient n
Power
y = f (g ( x )) Composite → Chain Rule y = f (u) f x y =k ( )
y = ln ( f ( x ) )
Exponential Natural Logarithm
y = sin ( f ( x ) )
y = cos ( f ( x ) )
Trigonometric
y = tan ( f ( x ) ) y = sin−1 ( f ( x ) ) y = cos−1 ( f ( x ) ) y = tan−1 ( f ( x ) )
Inverse Trionometric
DERIVATIVE
dy dx dy dx dy dx dy dx
=0 = nxn−1
= af ' ( x ) bg' ( x )
dv du +v dx dx du dv v −u dy dx dx = 2 dx v n−1 dy = n f ( x ) f ' ( x ) dx u = g(x) , y = f (u ) du dy = g' ( x ) = f ' (u ) dx du
dy dx dy dx dy dx dy dx
dy dx dy dx dy dx dy dx
=u
dy du du dx du = f ' (u ) dx f x = k ( ) lnk f ' ( x ) =
=
f ' (x) f (x)
= cos ( f ( x ) ) f ' ( x )
= − sin ( f ( x ) ) f ' ( x ) = sec 2 ( f ( x ) ) f ' ( x ) =
dy =− dx
f '(x) 1 − ( f ( x ))
2
f ' (x) 1 − ( f ( x ))
f '(x) dy = dx 1 + ( f ( x ) )2
2
IMPLICIT DIFFERENTIATION
Apply appropriate rule and differentiate with respect to x Equation cannot be expresses explicitly
STEPS TO DIFFERENTIATE AN EQUATION IMPLICITLY
differentiate both side with respect to x
differentiate every term with respect to x
đ?‘‘đ?‘Ś
đ?‘‘đ?‘Ś
đ?‘‘đ?‘Ś
separate terms with đ?‘‘đ?‘Ľ from non đ?‘‘đ?‘Ľ terms (LHS : đ?‘‘đ?‘Ľ term) đ?‘‘đ?‘Ś
factorize đ?‘‘đ?‘Ľ
đ?‘‘đ?‘Ś
đ?‘‘đ?‘Ś
simplify for đ?‘‘đ?‘Ľ ( đ?‘‘đ?‘Ľ as subject )