Differentiation 81

Example 4

Using your calculator, find an equation of the tangent to the curve f(x)=x^2 − 3 x+2at

x=5.

Find the slope of the tangent to the curve atx=5 by enteringd(x^2 − 3 x+2, x)|x=5.

The result is 7. Compute f(5)=12. Thus, the point (5, 12) is on the curve of f(x).

An equation of the line whose slopem=7 and passing through the point (5, 12) is

y− 12 =7(x−5).

TIP • Remember that d
dx
lnx=

1

x

and

∫

lnxdx=xlnx−x+c. The integral formula is

not usually tested in the BC exam.

The Chain Rule

If y = f(u) andu= g(x) are differentiable functions ofu and x respectively, then

d

dx

[f(g(x))]= f′(g(x))·g′(x)ordy

dx

=

dy

du

·

du

dx

.

Example 1

Ify=(3x−5)^10 , find

dy

dx

.

Using the chain rule, letu= 3 x−5 and thus,y=u^10. Then,

dy

du

= 10 u^9 and

du

dx

=3.

Since

dy

dx

=

dy

du

·

du

dx

,

dy

dx

=

(

10 u^9

)

(3)=10(3x−5)^9 (3)=30(3x−5)^9. Or you can use your

calculator and enterd((3x−5)^10 , x) and obtain the same result.

Example 2

Iff(x)= 5 x

√

25 −x^2 , findf′(x).

Rewrite f(x)= 5 x

√

25 −x^2 as f(x)= 5 x(25−x^2 )

(^12)
.Using the product rule, f′(x)=
(25−x^2 )
12 d
dx
(5x)+(5x)
d
dx
(25−x^2 )
(^12)
=5(25−x^2 )
(^12)
+(5x)
d
dx
(25−x^2 )
(^12)
.
To find
d
dx
(25−x^2 )
(^12)
, use the chain rule and letu= 25 −x^2.
Thus,
d
dx
(25−x^2 )
(^12)

1

2

(25−x^2 )−

(^12)
(− 2 x)=
−x
(25−x^2 )
12. Substituting this quantity back
into f′(x), you have f′(x)=5(25−x^2 )
(^12)
+(5x)
(
−x
(25−x^2 )
(^12)
)

5(25−x^2 )− 5 x^2
(25−x^2 )

12 =

125 − 10 x^2

(25−x^2 )

12 .Or you can use your calculator and enterd(5x

√

25 −x^2 , x) and obtain the

same result.