IMPLICIT DIFFERENTIATION

Transcription

1 Mathematics Revision Guies Implicit Differentiation Page 1 of Author: Mark Kulowski MK HOME TUITION Mathematics Revision Guies Level: AS / A Level AQA : C4 Eecel: C4 OCR: C4 OCR MEI: C3 IMPLICIT DIFFERENTIATION Version : 4 Date:

2 Mathematics Revision Guies Implicit Differentiation Page of Author: Mark Kulowski IMPLICIT DIFFERENTIATION A function involving an y is sai to be eplicit if y is clearly efine as a function of alone y = + 3 is an eample of an eplicit function; y is on the LHS, an the RHS is epresse fully in terms of y + 3y = 3 an + y = 16, on the other han, are implicit functions involving an y Sometimes an implicit equation in an y can be rearrange into an eplicit one by separating the variables Eample (1): Differentiate y = 3 with respect to The equation can be rearrange as y 3, an from there we can irectly work out = 3 Eample (): Differentiate - y = 7 with respect to Again we can separate the variables: - y = 7 7 = y y 7 y 7 Having obtaine y eplicitly in terms of, we can ifferentiate using the chain rule Let u = 7 u 1 an u u =, = 7 When an equation in an y is given implicitly an the variables cannot be separate, we fin by ifferentiating each term with respect to We must also remember that f ( y)) ( f ( y)) ( by the chain rule Eample (3): Differentiate + y with respect to Here there is no right-han sie to the epression, an so we cannot separate the variables as in the first two eamples, an so the result will be given in terms of, y an ( y ) ( ) ( y ) ( y ) 4 y Note the use of the chain rule in when ifferentiating y with respect to ( y ) ( y ) = y

3 Mathematics Revision Guies Implicit Differentiation Page 3 of Author: Mark Kulowski Eample (4): Differentiate y with respect to We nee to use the chain rule an the prouct rule here ( y) () ( y) () ( y) ( y) y Eample (): Differentiate y with respect to Again, we nee to use the chain rule an the prouct rule ( y ) ( ) ( y ) ( ) ( y ) ( y ) y y Eample (6): Fin in terms of an y if y + 6 = This time, there is a little more algebra involve, where we nee to bring the term over to the LHS Differentiating each term with respect to we have ( y ) (6) ( ) 3 y 6 y 6 y Eample (7): Fin the graient of the tangent to the curve - y = 7 (from Eample ) at the point (, 3) by implicit methos We coul work out the graient at = by substituting into the eplicit erivative = 7 (from E ) = 9 6 The question asks for implicit methos, so rewriting as - y - 7 = 0 an ifferentiating term by term will prouce an epression for in terms of an y ( y 7) ( ) (y ) 4y Rearranging an finally 4y = 0 to make the subject, we have y 4y Substituting = an y = 3 gives a graient of 6 at that point

4 Mathematics Revision Guies Implicit Differentiation Page 4 of Author: Mark Kulowski Eample (8): Fin in terms of an y if y = 8y This eample is a little trickier, since here we nee to use the prouct rule to ifferentiate y an 8y ( y ) ( ) ( y ) ( ) ( y ) ( y ) y y ( 8y) (8) ( y) (8) ( y) ( 8y) 8y 8 Having worke those two proucts out, we evaluate the whole erivative term by term an rearrange to make the subject ( y ) ( 0) (8y) y y 8y 8 y 8y 8 y y 8 8y y y (8 y) ( y 4)

5 Mathematics Revision Guies Implicit Differentiation Page of Author: Mark Kulowski Eamination questions usually inclue revision of AS-level techniques such as maima, minima, tangents an normals Eample (9): (Etension to Eample (8)) Fin the equation of i) the tangent to the curve y = 8y at the point (-4, 0); ii) the normal to the same curve at the point (,) Part i) Substitute (, y) = (-4, 0) into the epression y(8 y) ( y 4) point The graient at (-4, 0) is obtaine in the previous question to obtain the graient of the tangent at that ( 8)( 4) The equation of the tangent to the curve, in the form y = m + c, is 3 0 ( 4) 3 y 3 8 y (In the form a + by + c = 0, the equation is 3y = y + 0 = 0) Part ii) Substitute (, y) = (, ) into the epression for in i) to obtain the graient of the tangent at that point This graient is (8 ) 10 4( 4) 4, so the graient of the normal at (, ) is The equation of the normal to the curve, in the form y = m + c, is ( ) y 4 4 y y (In the form a + by + c = 0, the equation is y = y - 9 = 0)