Edexcel past paper questions - KUMAR'S MATHS REVISION

Edexcel past paper questions

Core Mathematics 4

Implicit Differentiation

C4 Implicit Differentiation

Edited by: K V Kumaran Email: kvkumaran@

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Implicit Differentiation 1. Introduction Normally, when differentiating, it is dealing with `explicit functions' of x, where a value of y is defined only in terms of x. However, some functions cannot be rearranged into this form, and we cannot express y solely in terms of x, therefore, we say y is given implicitly by x. Even so, given a value of x, a value for y can still be found, after a bit of work.

e.g.

y = 2x2 - 3x + 4 is expressed explicitly in terms of x.

x2 + y2 - 6x + 2y = 0

is expressed implicitly.

In general, to differentiate an implicit function to find, we differentiate both sides of the equation with respect to x. This allows you differentiate without making y the subject first.

In order to differentiate both sides of the equation, you will end up having to differentiate a term in y with respect to x. To do this, you follow the rule

()

=

-1

Basically

this

means

you

differentiate

y

as

it

were

x,

then

add

on

onto the end

of it. Effectively this is what you do when differentiating explicitly, but the y

differentiates to 1, leaving only dy/dx.

C4 Implicit Differentiation

Page 2

Another problem is differentiating something like 4xy2. This can be implicitly differentiated using the product rule, taking u = 4x and v = y2. Remembering to leave dy/dx after each y term that has been differentiated, it becomes

[()()]

=

()()

+

()()

After differentiating each term in turn, you then need to rearrange the equation

to

find

on

its

own.

nb: the function y=ax differentiates to axln(a), and this can be shown through implicit differentiation by taking logs of both sides.

C4 Implicit Differentiation

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C4 Implicit Differentiation

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C4 Implicit Differentiation

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