Math 180: Calculus I
Fall 2014
October 7
TA: Brian Powers
We may use the following derivative rules now:
d x
b = bx ln(b)
dx
d
1
ln x =
dx
x
d
u0 (x)
ln |u(x)| =
dx
u(x)
d
1
logb x =
dx
x ln b
And the technique of logarithmic differentiation: take a log of both sides of the equation, then take the
derivative using implicit differentaition to solve for f 0 (x).
1. Find the following derivatives
(a)
d
2
dx (x
ln x)
SOLUTION:
= 2x ln x + x2
(b)
1
x
d 3 x
dx x 3
SOLUTION:
= 3x2 3x + x3 (3x ln 3)
(c)
d
dx (ln | sin x|)
SOLUTION:
=
(d)
cos x
= cot x
sin x
d
dx
ln(10x )
SOLUTION:
=
10x ln 10
= ln 10
10x
Or you may use hte fact that
ln(10x ) = x ln 10
and simply take that derivative (ln 10 is just a constant).
(e)
d
dx (ln(ln x))
SOLUTION:
=
1/x
1
=
ln x
x ln x
2. Find the derivatives
(a) s(t) = cos(2t )
SOLUTION:
s0 (t) = − sin(2t )(2t ln t)
(b) f (x) = ln (x3 + 1)π
SOLUTION:
f 0 (x) =
π(x3 + 1)π−1 (3x2 )
(x3 + 1)π
11-1
11-2
TA Discussion 11: October 7
√
3. Evaluate the derivative of h(x) = x x at x = 4.
SOLUTION: Using logarithmic differentiation, we take the natural log of both sides first
√ ln h(x) = ln x x
√
ln h(x) = x ln x
by log property
0
√ 1
h (x)
1
= √ ln x + x
take derivative of both sides
h(x)
x
2 x
1
ln x
√ +√
h0 (x) = h(x)
solve for h0 (x)
2 x
x
√
ln x
1
x
0
√ +√
h (x) = x
substitute in h(x)
2 x
x
√
1
ln 4
√ +√
plug in 4
h0 (4) = 4 4
2 4
4
ln 4 1
= 16
+
4
2
= 4 ln 4 + 8
4. Find the horizontal tangent line equation for y = xln x
SOLUTION: Using logarithmic differentiation,
ln y = ln xln x
ln y = (ln x)(ln x) = (ln x)2
by log property
0
1
y
= 2(ln x)
y
x
2 ln x
y0 = y
x
0
ln x 2 ln x
y =x
x
take derivative of both sides
solve for y 0
substitute in y
Because x = 0 is not in the domain, the only way this derivative can be zero is for ln x = 0. This
happens when x = 1 (this is true for all logs, no matter what base). Plugging in 1 for x we get
y = (1)ln 1 = 10 = 1
So the equation of the horizontal tangent line is
y=1
5. Use logarithmic differentiation to find the derivative of
x8 cos3 x
f (x) = √
x−1
TA Discussion 11: October 7
11-3
SOLUTION:
ln f (x) = ln
x8 cos3 x
√
x−1
take log of both sides
√
ln f (x) = ln(x8 ) + ln(cos3 x) − ln( x − 1)
by log properties
1
ln f (x) = 8 ln x + 3 ln(cos x) − ln(x − 1)
by log properties
2
f 0 (x)
8
sin x
1 1
= −3
−
take derivative of both sides
f (x)
x
cos x 2 x − 1
8
1
= − 3 tan x +
x
2 − 2x
8
1
0
f (x) = f (x)
− 3 tan x +
solve for f 0 (x)
x
2 − 2x
8
x cos3 x
8
1
= √
− 3 tan x +
substitute back f (x)
x
2 − 2x
x−1
6. Find the derivative y 0 of
y = (x2 + 1)x
using two methods:
(1) Use the fact that
bx = ex ln b
(2) Use logarithmic differentiation.
SOLUTION:
Method (1):
y = ex ln(x
So
0
y =e
x ln(x2 +1)
2
+1)
2x
2x2
2
2
x
2
ln(x + 1) + x 2
= (x + 1) ln(x + 1) + 2
x +1
x +1
Method (2):
ln y = ln (x2 + 1)x
= x ln(x2 + 1)
x
y0
= ln(x2 + 1) + 2
(2x)
y
x +1
2x2
0
2
y = y ln(x + 1) + 2
x +1
2x2
2
x
2
= (x + 1) ln(x + 1) + 2
x +1