MA 3972-MA-Book May 9, 2018 10:9
Differentiation 121
Example 8
y=ln(lnx)
Letu=lnx. Then
dy
dx
=
1
lnx
du
dx
=
1
lnx
(
1
x
)
=
1
xlnx
.
Example 9
y=log 5 (2x+1)
Letu= 2 x+1. Then
dy
dx
=
1
(2x+1) ln 5
du
dx
=
1
(2x+1) ln 5
·(2)=
2
(2x+1) ln 5
.
Example 10
Write an equation of the line tangent to the curve ofy=exatx=1.
The slope of the tangent to the curvey=exatx=1 is equivalent to the value of the
derivative ofy=exevaluated atx=1. Using your calculator, enterd(e∧(x), x)|x=1 and
obtaine. Thus,m=e, the slope of the tangent to the curve atx=1. Atx=1,y=e^1 =e, and
thus, the point on the curve is (1,e). Therefore, the equation of the tangent isy−e=e(x−1)
ory=ex. (See Figure 7.2-1.)
[–1, 3] by [–2, 8]
Figure 7.2-1
TIP • Never leave a multiple-choice question blank. There is no penalty for incorrect answers.
7.3 Implicit Differentiation
Main Concept:Procedure for Implicit Differentiation
Procedure for Implicit Differentiation
Given an equation containing the variablesxandyfor which you cannot easily solve fory
STRATEGY
in terms ofx, you can find
dy
dx
by doing the following:
Steps
- Differentiate each term of the equation with respect tox.
- Move all terms containing
dy
dx
to the left side of the equation and all other terms to the
right side. - Factor out
dy
dx
on the left side of the equation. - Solve for
dy
dx