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Differential Calculus
Tangent Line And Normal Line
The Tangent Line
To find the derivative, one can use its definition which is f ' ( x ) 

lim

f ( x  x )  f ( x )

...

*Derivative - Function that tells the tangent line of the curve; slope
This would be further understood with some examples
Find an equation for the tangent line to the parabola y  x 2  1 at the point (2,3)

lim

f ( x  x )  f ( x )
x  0
x
lim f (x  2)  f (2)
f ' ( x) 
x  0
x
2
lim ((x  2)  1)  (2 2  1)
f ' ( x) 
x  0
x
2
lim ((x)  4x  4  1)  (4  1)
f ' ( x) 
x  0
x
lim (x) 2  4x  4  1  4  1
f ' ( x) 
x  0
x
lim (x) 2  4x
f ' ( x) 
x  0
x
lim x(x  4)
f ' ( x) 
x  0
x
lim
f ' ( x) 
x  4
x  0
f ' ( x)  0  4
f ' ( x)  4
f ' ( x) 

Write the definition of derivative
Substitute the x-coordinate, 2, to x
Substitute to the given equation
Expand the square binomial and simplify
Distribute the negative sign
Add common factors
Factor out x
Cancel out x
Find the limit
Simplify

Since the tangent line of any equation is the slope of that same equation, the slope of the
equation y  x 2  1 is 4, which is also its derivative
...
What is the equation of the normal line?

*derivative=4

1

*m=  4

X 2
 3
4 4
 X  14
Y
4

Y  Y1  m( X  X 1 )
1
Y  3   ( X  2)
4
X 2
Y 3   
4 4
The equation of the normal line is Y 

Y 

 X  14
4

2
The curve Y  7  4 x  x passes through the point (-1, 5)
...
What is the equation of the normal

line?

y   x 2 ( x  1)
lim f (x  x)  f ( x)
m
x  0
x
lim f (x  2)  f (2)
m
x  0
x
m
m

lim

((x  2) 2 (x  2  1))  (2 2 (2  1))
x  0
x
2
lim (((x)  4x  4)(x  3))  (4(3))
x  0

x

m

lim

 ((x) 3  4(x) 2  4x  3(x) 2  12x  12)  12
x  0
x
3
lim  ((x)  7(x) 2  16x  12)  12
m
x  0
x
3
lim  (x)  7(x) 2  16x  12  12
m
x  0
x
lim
m
 (x) 2  7 x  16
x  0
m  (0) 2  7(0)  16
m  16

TANGENT

m  16
Y  Y1  m( X  X 1 )
1
Y   16( X  2)
2
1
Y   16 X  32
2
1
Y  16 X  32 
2
65
Y  16 X 
2

NORMAL

1
16
Y  Y1  m( X  X 1 )
1 1
Y   ( X  2)
2 16
1 X 2
Y  
2 16 16
X 2 1
Y  
16 16 2
X 6
Y
16
m



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