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082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

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11.3 Parametric, implicit and logarithmic differentiation 395

Solution (a) We make a substitution and letz = y 4 so that the problem becomes that of finding

dz

. Now, usingthe chain rule,

dx

dz

dx = dz

dy × dy

dx

Ifz =y 4 then dz

dy = 4y3 and so

dz

dx = 4y3dy dx

We conclude that

d

dx (y4 ) = 4y 3dy

dx

(b) Wemakeasubstitutionandletz =y −3 sothattheproblembecomesthatoffinding

dz

dxİfz =y−3 then dz

dy = −3y−4 and so

d

dx (y−3 ) = dz

dx = dz

dy × dy

dx = −3y−4dy dx

Example11.11 Find d dt (lny).

Solution Weletz = lnysothattheproblembecomesthatoffinding dz

dt .Ifz=lnythendz dy = 1 y

and so usingthe chain rule

dz

dt = dz

dy × dy

dt = 1 dy

y dt

weconclude that

d

dt (lny) = 1 dy

y dt

Examples11.10and 11.11illustrate the generalformula:

d

dx (f(y))=df dy × dy

dx .

This issimplythe chain ruleexpressed inaslightly different form.

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