PracticeMidterm2Solutions - Practice Midterm 2 Solutions 1...

Practice Midterm 2 Solutions
1.
a) Using the chain rule
dy
dx
= e
x 3 +1
· 3 x
2
b) Using the change of base formula,
y =
ln cos x
ln x
so from the quotient and chain rules,
dy
dx
=
ln x · 1
cos x
( - sin x ) - 1
x
ln cos x
(ln x )
2
c) Applying
d
dx
to each side and using the product and chain rules
2 y
dy
dx
+ 6 x
dy
dx
+ 6 y + 2 x = 0 ⇒
dy
dx
=
- 3 y - x
y + 3 x
2. Let ‘ be the distance from the origin to the point q . By the Pythagorean
theorem, ‘
2
= x
2
+ y
2
. Applying
d
dt
to each side and using the chain rule
gives
2 ‘
d‘
dt
= 2 x
dx
dt
+ 2 y
dy
dt
⇒
d‘
dt
=
x
‘
·
dx
dt
+
y
‘
·
dy
dt
Since y = x 2 , by the chain rule
dy
dt
= 2 x
dx
dt
, so
d‘
dt
=
( x + 2 x 3 )
√
x 2 + x 4
·
dx
dt
= sign ( x )
1 + 2 x
√
1 + x 2
·
dx
dt
3. Approximating sin by its tangent line at 0,
sin 1
◦
= sin
π
180
≈ sin 0 + (cos 0)
π
180
- 0 =
π
180
4.
a) Method 1 Recognizing the limit as a derivative,
lim
b → 2
b 2010 - 2 2010
b - 2
=
d
db
(
b
2010
)
b =2
= 2010 · 2
2009

Method 2 Factoring b 2010 out of the numerator and b out of the denominator