Post on 14-Jan-2016
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)