MA 3972-MA-Book May 9, 2018 10:9Differentiation 115Example 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 AB 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)or
dy
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).
Rewritef(x)= 5 x√
25 −x^2 as f(x)= 5 x(25−x^2 )^1 /^2 .Using the product rule,
f′(x)=(25−x^2 )^1 /^2
d
dx
(5x)+(5x)
d
dx
(25−x^2 )^1 /^2 =5(25−x^2 )^1 /^2 +(5x)
d
dx
(25−x^2 )^1 /^2.To find
d
dx
(25−x^2 )^1 /^2 , use the chain rule and letu= 25 −x^2.Thus,
d
dx(25−x^2 )^1 /^2 =1
2
(25−x^2 )−^1 /^2 (− 2 x)=
−x
(25−x^2 )^1 /^2. Substituting this quantity back
into f′(x), you have f′(x)=5(25−x^2 )^1 /^2 +(5x)(
−x
(25−x^2 )^1 /^2)
=
5(25−x^2 )− 5 x^2
(25−x^2 )^1 /^2=
125 − 10 x^2
(25−x^2 )^1 /^2
.Or you can use your calculator and enterd(5x√
25 −x^2 , x) and obtain thesame result.