The document discusses differentiability and implicit differentiation. It defines differentiability as a function being smooth and continuous, with the left and right hand limits of the derivative being equal. An example of a non-differentiable function is f(x) = 1/x at x=1, as the left and right limits are not equal. Implicit differentiation is introduced as a way to find derivatives of implicitly defined functions using the relationship df/dx = df/dy * dy/dx. Examples are worked through, such as finding the derivative of x=y^2 and the equation of the tangent line to an implicitly defined function.

1. Differentiability

2. Differentiability
A function is differentiable at a point if the curve is smooth continuous
i.e. lim f   x   lim f   x 
x a x a

3. Differentiability
A function is differentiable at a point if the curve is smooth continuous
i.e. lim f   x   lim f   x 
x a x a
y
y  x 1
1
1 x

4. Differentiability
A function is differentiable at a point if the curve is smooth continuous
i.e. lim f   x   lim f   x 
x a x a
y
y  x 1
1
1 x
not differentiable at x = 1

5. Differentiability
A function is differentiable at a point if the curve is smooth continuous
i.e. lim f   x   lim f   x 
x a x a
y
y  x 1
1
1 x
not differentiable at x = 1
lim f   x   1

x1

6. Differentiability
A function is differentiable at a point if the curve is smooth continuous
i.e. lim f   x   lim f   x 
x a x a
y
y  x 1
1
1 x
not differentiable at x = 1
lim f   x   1

lim f   x   1

x1 x1

7. Differentiability
A function is differentiable at a point if the curve is smooth continuous
i.e. lim f   x   lim f   x 
x a x a
y y y  x2
y  x 1
1
1 x x
not differentiable at x = 1
lim f   x   1

lim f   x   1

x1 x1

8. Differentiability
A function is differentiable at a point if the curve is smooth continuous
i.e. lim f   x   lim f   x 
x a x a
y y y  x2
y  x 1
1
1 x 1 x
not differentiable at x = 1
lim f   x   1

lim f   x   1

x1 x1

9. Differentiability
A function is differentiable at a point if the curve is smooth continuous
i.e. lim f   x   lim f   x 
x a x a
y y y  x2
y  x 1
1
1 x 1 x
not differentiable at x = 1 differentiable at x = 1
lim f   x   1

lim f   x   1

x1 x1

10. Implicit Differentiation

11. Implicit Differentiation
df df dy
 
dx dy dx

12. Implicit Differentiation
df df dy
 
dx dy dx
e.g. (i) x  y 2

13. Implicit Differentiation
df df dy
 
dx dy dx
e.g. (i) x  y 2
 x   y 
d d 2
dx dx

14. Implicit Differentiation
df df dy
 
dx dy dx
e.g. (i) x  y 2
 x   y 
d d 2
dx dx
d 2
dx
 y   dy  y   dx
d 2 dy

15. Implicit Differentiation
df df dy
 
dx dy dx
e.g. (i) x  y 2
 x   y 
d d 2
dx dx
dy
1 2y
dx
d 2
dx
 y   dy  y   dx
d 2 dy

16. Implicit Differentiation
df df dy
 
dx dy dx
e.g. (i) x  y 2
 x   y 
d d 2
dx dx
dy
1 2y
dx
dy 1

dx 2 y
d 2
dx
 y   dy  y   dx
d 2 dy

17. Implicit Differentiation
df df dy
 
dx dy dx
e.g. (i) x  y 2 (ii)
dx
x y 
d 2 3
 x   y 
d d 2
dx dx
dy
1 2y
dx
dy 1

dx 2 y
d 2
dx
 y   dy  y   dx
d 2 dy

18. Implicit Differentiation
df df dy
 
dx dy dx
e.g. (i) x  y 2 (ii)
dx
x y 
d 2 3
 x   y 
d d 2
  x2   3 y 2    y3   2 x 
dx dx dy
 
dy  dx 
1 2y
dx
dy 1

dx 2 y
d 2
dx
 y   dy  y   dx
d 2 dy

19. Implicit Differentiation
df df dy
 
dx dy dx
e.g. (i) x  y 2 (ii)
dx
x y 
d 2 3
 x   y 
d d 2
  x2   3 y 2    y3   2 x 
dx dx dy
 
dy  dx 
1 2y 2 2 dy
dx  3x y  2 xy 3
dy 1 dx

dx 2 y
d 2
dx
 y   dy  y   dx
d 2 dy

20.  iii  Find the equation of the tangent to x 2  y 2  9 at the point 1, 2 2 

21.  iii  Find the equation of the tangent to x 2  y 2  9 at the point 1, 2 2 
x2  y 2  9

22.  iii  Find the equation of the tangent to x 2  y 2  9 at the point 1, 2 2 
x2  y 2  9
dy
2x  2 y  0
dx

23.  iii  Find the equation of the tangent to x 2  y 2  9 at the point 1, 2 2 
x2  y 2  9
dy
2x  2 y  0
dx
dy
2 y  2 x
dx

24.  iii  Find the equation of the tangent to x 2  y 2  9 at the point 1, 2 2 
x2  y 2  9
dy
2x  2 y  0
dx
dy
2 y  2 x
dx
dy x

dx y

25.  iii  Find the equation of the tangent to x 2  y 2  9 at the point 1, 2 2 
x2  y 2  9
dy
2x  2 y  0
dx
dy
2 y  2 x
dx
dy x

dx y
 
dy
at 1, 2 2 ,  
dx
1
2 2

26.  iii  Find the equation of the tangent to x 2  y 2  9 at the point 1, 2 2 
x2  y 2  9
dy
2x  2 y  0
dx
dy
2 y  2 x
dx
dy x

dx y
 
dy
at 1, 2 2 ,  
dx
1
2 2
1
 required slope  
2 2

27.  iii  Find the equation of the tangent to x 2  y 2  9 at the point 1, 2 2 
x2  y 2  9 1
dy y2 2    x  1
2x  2 y  0 2 2
dx
dy
2 y  2 x
dx
dy x

dx y
 
dy
at 1, 2 2 ,  
dx
1
2 2
1
 required slope  
2 2

28.  iii  Find the equation of the tangent to x 2  y 2  9 at the point 1, 2 2 
x2  y 2  9 1
dy y2 2    x  1
2x  2 y  0 2 2
dx
2 2y  8  x 1
dy
2 y  2 x
dx
dy x

dx y
 
dy
at 1, 2 2 ,  
dx
1
2 2
1
 required slope  
2 2

29.  iii  Find the equation of the tangent to x 2  y 2  9 at the point 1, 2 2 
x2  y 2  9 1
dy y2 2    x  1
2x  2 y  0 2 2
dx
2 2y  8  x 1
dy
2 y  2 x
dx x  2 2y  9  0
dy x

dx y
 
dy
at 1, 2 2 ,  
dx
1
2 2
1
 required slope  
2 2

30.  iii  Find the equation of the tangent to x 2  y 2  9 at the point 1, 2 2 
x2  y 2  9 1
dy y2 2    x  1
2x  2 y  0 2 2
dx
2 2y  8  x 1
dy
2 y  2 x
dx x  2 2y  9  0
dy x

dx y
 
dy
at 1, 2 2 ,  
dx
1
2 2
1 Exercise 7K; 1acegi, 2bdfh, 3a,
 required slope  
2 2 4a, 7, 8