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# Maidstone Grammar School for Girls ~ A level Mathematics (C4)

By Christopher Adams,2014-01-07 06:40
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Maidstone Grammar School for Girls ~ A level Mathematics (C4) Implicit Differentiation L.O.: To find derivatives for functions defined implicitly. To use implicit differentiation to be able to find the gradient of a curve, the equation of a tangent and to locate turning points. Functions can be defined explicitly or implicitly:- Explicit functions have the ..

Maidstone Grammar School for Girls ~ A level Mathematics (C4)

Implicit Differentiation

L.O.: To find derivatives for functions defined implicitly.

To use implicit differentiation to be able to find the gradient of a curve, the equation of a tangent

and to locate turning points.

Functions can be defined explicitly or implicitly:-

Explicit functions have the form y = f(x) Implicit functions do not have y as the

SUBJECT

222 xy？，9 yx，？9

2325x xy36 yxe

33 yx，？5sin(2) xyxy？，

22Not all functions can be expressed in the form y = f(x), whilst other functions (such as ) xy？，9are simpler written implicitly.

Introductory example:

22Find the gradient of the curve at the point . xy？，92,5;；

Solution:

22We can differentiate term by term:- xy？，9

22 xy？，9

Easy to differentiate with How do we differentiate Easy to differentiate 2 with respect to x? with respect to x. respect to x. y

2We can use the chain rule to differentiate with respect to x:- y

dddydy22yyy，(，2 ;；;；dxdydxdx

22Therefore, when you differentiate , you get: xy？，9

dy 220xy？，dx

i.e.

dySo dx

dySo at the point , 2,5;；dx

Note:

d (sin)ydx

d3 = (23)xydx

dy ()edx

dy23Example: Find an expression for on the curve xxy？？，22dx

Differentiating term by term:

dySo = dx

32Example 2: Find for the implicit function yxxyy？，25

Applications to tangents, normals and turning points

22Example 1: A curve has equation . xxyy？？，311

a) Find the equation of the tangent to the curve at the point (1, 2).

b) Show that the normal to the curve at (1, 2) goes through the point (9, 9).

Solution:

a) Differentiating the terms in the equation with respect to x gives:

Therefore at the point (1, 2) the gradient is:

The equation of a tangent is . yymxx？，？()11

The gradient m =

So the equation is:

Rearranging:

b) The gradient of the normal at (1, 2) is:

The equation of the normal is yymxx？，？()11

Therefore:

So the equation of the normal is

When x = 9, we find that y =

This is as required.

22Example 2: Find the coordinates of all the stationary points on the curve xyxy？？，3

Plenary: Examination style question

22a) Find the gradient to the curve at the point (2, 0). xxyyxy？？，？？332

b) Find the equation of the tangent to the curve at this point.

Correct the “solution” to this question given below:

a) Differentiating the terms in the equation with respect to x:

dydydy 232132xy？？，？？dxdxdx

dy ；，？232yxdx

dyx32 ；，dxy2

So at the point (2, 0):

dy3221？(？ ；，，，？1dx200(

b) The equation of a tangent is

y = mx + c.

The gradient, m, is 1.

So yxc，？？1

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