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

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396 Chapter 11 Techniques of differentiation

Example11.12 Find d dx (y).

Solution We have

d dy

(y) =

dx dy × dy

dx = dy

dx

Example11.13 Given

find dy

dx .

x 3 +y=1+y 3

Solution Consider differentiation of the l.h.s.w.r.t.x.

d

dx (x3 +y)= d dx (x3 )+ dy

dx =3x2 + dy

dx

Now consider differentiation of the r.h.s.w.r.t.x.

d

dx (1+y3 )= d dx (1) + d dx (y3 )= d dx (y3 )

d

We note from the formulafollowing Example 11.11 that

dx (y3 ) = 3y 2dy . So finally,

dx

3x 2 + dy

dx = 3y2dy dx

from which

dy

dx = 3x2

3y 2 −1

Note that dy isexpressed intermsofxandy.

dx

Example11.14 Find dy

dx given

(a) lny=y−x 2

(b) x 2 y 3 −e y = e 2x

Solution (a) Differentiating the given equation w.r.t.xyields

1dy

y dx = dy

dx −2x

from which

dy

dx = 2xy

y −1

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