Try out the starter

Change them into Indices…

…then differentitate!!!

4 3x

2

3

x

xx

23 x

43x

23x

36 x

41

4

3 x

21

2

3x

121

6 xx

21

31

2 xx

221

62

1 xx

23

32

3

1 xx

xx6

xx

23

21

5 xx

x

x 5

21

21

5 xx

23

21

2

5

2

1 xx

C2 Differentiation – Tangents and Normals

• How to find the gradient of a line• How to find the line equation of a tangent• How to find the line equation of a normal• All of the above, but with negative and

fractional powers

C2 Differentiation2xy

xdx

dy2 Tells us a gradient…

But a gradient of what?

2xy 6m

xdx

dy2 Tells us the gradient of the

tangent at the point x = … 3x

2xy

4m

xdx

dy2 Tells us the gradient of the

tangent at the point x = … 2x

2xy

2m

xdx

dy2 Tells us the gradient of the

tangent at the point x = … 1x

2xy

0m

xdx

dy2 Tells us the gradient of the

tangent at the point x = … 0x

2xy

2m

xdx

dy2 Tells us the gradient of the

tangent at the point x = … 1x

2xy

4m

xdx

dy2 Tells us the gradient of the

tangent at the point x = … 2x

2xy

6m

xdx

dy2 Tells us the gradient of the

tangent at the point x = … 3x

dy/dx is the Gradient of a tangent to the curve!

Process of finding a tangent

Step 1) Find a point for the line. It can be a given point or you have to find it by sub-ing in a x-value in a given equation

Step 2) Find gradient via differentiation

Step 3) Use m & your point with the line equation 11 xxmyy

Finish off the exam questions

Last lesson you started the first parts of an exam questions, finding the first derivative.

These are normally followed by finding the tangent

4

83x

xy 483 xxy

5321 xdx

dy

5321 xdx

dy

51321 dx

dy

31m

418

31 y

12y

13112 xy

Process of finding a normal

Step 1) Find a point for the line. It can be a given point or you have to find it by sub-ing in a x-value in a given equation

Step 2) Find a gradient via differentiation, then find the perp. gradient!

Step 3) Use m & your point with the line equation 11 xxmyy

1

1

mmperp

Finish off the exam question

Finish off the exam question and find the line equation of the normal!

x

xx 3

x

x

x

x 213

212 xx

212 xx 2

3

212 xx

dx

dy

23

212 xx

dx

dy

5.1

212 xxy

21

11 2 2

2

31 m

3

22 m 1

3

22 xy

Try out the exam question on back

Here’s a brand new exam question for you!

2

3

2

1

4 xxy

dx

dy4

2

1

121 x

2

3 123 x2

221 x 2

223 x

2

3

2

1

4 xxy

21

21

2

32 xx

dx

dy

2

3

2

1

4 xxy 31

21

2

32 xx

dx

dy

)0,4( 4x

21

21

42

342

dx

dy2

2

3

2

1

4 xxy 31

21

2

32 xx

dx

dy

2tan m 1tan normm

)( 11 xxmyy )4(2

10 xy 2

2

1 xy

Try out the exam questions on back

Independent Study

Misc. Exercise 8 p130 (solution p423)