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

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390 Chapter 11 Techniques of differentiation

Example11.3 Giveny =z 6 wherez =x 2 +1 find dy

dx .

Solution If y = z 6 and z = x 2 + 1, then y = (x 2 + 1) 6 . We recognize this as the composition

y(z(x)) (see Section 2.3.6). Now y = z 6 and so dy

dz = 6z5 . Alsoz = x 2 +1 and so

dz

= 2x. Usingthe chain rule,

dx

dy

dx = dy

dz × dz

dx =6z5 2x =12x(x 2 +1) 5

Example11.4 Ify = ln(3x 2 +5x +7), find dy

dx .

Solution Weuseasubstitutiontosimplifythegivenfunction:letz = 3x 2 +5x+7sothaty = lnz.

Since

dy

y=lnz then

dz = 1 z

Also

z=3x 2 +5x+7

Using the chain rulewefind

dy

dx = dy

dz × dz

dx

= 1 z ×(6x+5)

=

6x+5

3x 2 +5x+7

andso

dz

dx =6x+5

Note that in the previous answer the numerator is the derivative of the denominator.

This result is true more generally and can be applied when differentiating the natural

logarithm ofany function:

Wheny = ln f(x)then dy

dx = f ′ (x)

f(x)

Example11.5 Findy ′ when

(a) y =ln(x 5 +8) (b) y =ln(1−t) (c) y =8ln(2−3t)

(d) y =ln(1 +x)

(e) y =ln(1 +cosx)

Solution Ineachcase weapplythe previousrule.

(a) Ify = ln(x 5 +8),theny ′ =

5x4

x 5 +8 .

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