Download Fall 2004 Math 151 4 Inverse Functions 4.4 Derivatives of

The domain of f is x : x ≤ log2 3 . (Note: log2 3 ≈ 1.58).
Fall 2004 Math 151
4 Inverse Functions
4.4 Derivatives of Logarithmic
Functions
c
Mon, 25/Oct
2004,
Art Belmonte
• The requirements for the domain of f 0 are almost the same,
with the additional provision that the expression under the
square root sign not be zero, lest
we divide by zero.
Therefore, the domain of f 0 is x : x < log2 3 .
269/10
Summary
Let y = ln (sec x + tan x). Find y 0 and y 00 .
Four derivatives
Solution
As always, these may be combined with the chain rule.
Via the chain rule, we have
1
d
(ln x) =
dx
x
1
d
loga x =
dx
x ln a
1
d
(ln |x|) =
dx
x
d
a x = a x ln a
dx
y0 =
sec x tan x + sec2 x
(sec x + tan x) sec x
=
= sec x.
sec x + tan x
(sec x + tan x)
In turn, y 00 = sec x tan x.
269/20
Logarithmic Differentiation
r
Differentiate G(u) = ln
This technique makes it easier to differentiate a function that
involves combinations of products, powers, and/or quotients.
1. Take the natural logarithm of both sides of y = f (x).
Apply the properties of logarithms when needed.
Solution
2. Implicitly differentiate with respect to x.
Rewrite G as G(u) = ln
3. Solve for d y/d x, replacing y by f (x) in the final result.
differentiate.
Hand Examples
G 0 (u) =
269/6
√
G 0 (u) =
3 − 2x . Find f 0 (x) and state the domains of f and
3u + 2
3u − 2
1/2 !
. Use log props, then
1
(ln (3u + 2) − ln (3u − 2))
2
3
1
3
−
[Stop; this is fine.]
2 3u + 2 3u − 2
1 9u − 6 − 9u − 6
2
9u 2 − 4
6
[Or simplify to taste!]
4 − 9u 2
G(u) =
Let f (x) =
f 0.
3u + 2
.
3u − 2
G 0 (u) =
Solution
269/23
• Write f as f (x) = (3 − 2x )1/2 . Then
Differentiate y =
−1/2
1
2 x−1 ln 2
3 − 2x
.
−2 x ln 2 = − √
f (x) =
2
3 − 2x
0
• For the domain of f , we need the expression under the
square root sign to be nonnegative.
3 − 2x
3
ln 3
ln 3
ln 2
log2 3
Solution
The quotient rule gives
≥ 0
≥ 2x
≥
x ln 2
≥
x
≥
x
ln x
.
1 + x2
1
dy
dx
=
dy
dx
=
1 + x2
1
− (ln x) (2x)
x
2
1 + x2
1 + x 2 − 2x 2 ln x
2 .
x 1 + x2
269/29
Solution
Compute the derivative of f (t) = π −t .
Once again, apply logarithmic differentiation.
y ln x
1
y 0 ln x + y ·
x
x
y0
ln x −
y
Solution
This is the exponential function whose base is the constant π.
ln π
Accordingly, f 0 (t) = π −t ln π (−1) = − t via the chain rule.
π
=
x ln y
=
1 · ln y + x ·
=
ln y −
y0
=
y0
=
269/30
Find the derivative of g(x) =
1.6 x
+
x 1.6 .
1 0
y
y
y
x
y
ln y −
x
x
ln x −
y
x y ln y − y 2
x y ln x − x 2
Solution
269/63
We have the sum of an exponential function and a polynomial.
So dg/d x = 1.6 x ln 1.6 + 1.6x 0.6 .
Use logarithmic differentiation to find the derivative of
p
ex x 5 + 2
y=
2 .
(x + 1)4 x 2 + 3
269/42
Solution
Compute the derivative of y = (sin x)cos x .
We have
Solution
Unlike the preceding problem, this is neither an exponential
function nor a polynomial. (Both the base and the exponent vary.)
Accordingly, use logarithmic differentiation.
ln y
=
x+
1 dy
y dx
=
1+
dy
dx
=
1. Take the natural logarithm of y = (sin x)cos x .
ln y = cos x · ln (sin x)
1
2
ln x 5 + 2 − 4 ln (x + 1) − 2 ln x 2 + 3
5x 4
2 (2x)
4
−
− 2
5
x +1 x +3
2 x +2
!
p
ex x 5 + 2
4x
5x 4
4
−
−
.
2 1 +
x + 1 x2 + 3
2 x5 + 2
(x + 1)4 x 2 + 3
269/65
If g = f −1 is the inverse function of f (x) = 2x + ln x, find g 0 (2).
2. Implicitly differentiate with respect to x.
cos x
1 dy
= − sin x · ln (sin x) + cos x ·
y dx
sin x
Solution
By inspection f (1) = 2, whence g(2) = f −1 (2) = 1. Therefore,
3. Solve for d y/d x, replacing y by (sin x)cos x in the final result.
g 0 (2) =
dy
= (sin x)cos x (cos x cot x − sin x ln (sin x))
dx
1
1
1
= 0
= 1 f 0 (g(2))
f (1)
2+ x 269/54
269/67
Find y 0 if x y = y x .
Use the definition of derivative to prove that lim
=
1
.
3
x=1
ln (1 + x)
= 1.
x
x→0
2
Solution
%-------------------------------------------------% Stewart 269/42
%
syms x
y = sin(x) ˆ cos(x); pretty(y)
d
1 ln (1 + x) − ln 1
=
=
= 1.
lim
(ln(1 + x)) x −0
dx
1 + x x=0
x→0
x=0
cos(x)
sin(x)
yp = diff(y,x); pretty(yp)
MATLAB Examples
/
2\
cos(x) |
cos(x) |
sin(x)
|-sin(x) log(sin(x)) + -------|
\
sin(x) /
s269x25
Differentiate y = ln x 3 − x 2 . Plot y and y 0 on the same figure.
%
echo off; diary off
Solution
s269x45
We have y 0 =
3x 2 − 2x
3x − 2
.
=
3
2
x(x − 1)
x −x
Let f (x) =
x
. Find f 0 (e). Illustrate.
ln x
Stewart 269/25
6
Solution
function
derivative
4
Via diff and subs, MATLAB reveals that f 0 (e) = 0. That is, the
tangent line at x = e is horizontal.
2
0
−2
Stewart 269/45
2.95
−4
−6
2.9
−8
−1
0
x
1
2.85
2
y
−10
−2
2.8
%-------------------------------------------------% Stewart 269/25
%
x = linspace(-2, 2, 201); % Yes, 201 so we hit x = 0, 1.
y = log(abs(x.ˆ3 - x.ˆ2));
yp = (3.*x - 2) ./ (x .* (x - 1));
plot(x,y, x,yp,’r-.’); grid on
legend(’function’, ’derivative’, ’Location’, ’NorthWest’)
axis([-2 2 -10 6])
xlabel(’x’); title(’Stewart 269/25’)
% Painting the asymptotes x = 0 and x = 1.
% 1. Apply electronic white-out.
% 2. Now overspray.
hold on
plot([0 0], [-10 6], ’w’)
plot([1 1], [-10 6], ’w’)
plot([0 0], [-10 6], ’g--’)
plot([1 1], [-10 6], ’g--’)
%
2.75
2.7
2
2.5
3
x
3.5
4
%-------------------------------------------------% Stewart 269/45
%
syms x
f = x / log(x);
fp = diff(f,x); pretty(fp)
1
1
------ - ------log(x)
2
log(x)
echo off; diary off
fp at e = subs(fp, x, exp(1))
fp at e =
0
x = linspace(2, 4);
f = eval(vectorize(f));
plot(x,f); grid on
hold on; e = exp(1);
plot([2 4], [e e], ’r--’)
plot(e, e, ’go’, ’MarkerFaceColor’, ’g’, ...
’MarkerSize’, 7)
xlabel(’x’); ylabel(’y’)
title(’Stewart 269/45’)
%
s269x42 [269/42 revisited]
Compute the derivative of y = (sin x)cos x .
Solution
We use diff to verify the derivative we computed by hand.
echo off; diary off
3
s269x47
Stewart 269/49
0.6
function f(x) = ln(ln(x))
tangent line y = x/e − 1
Let f (x) = sin x + ln x. Compute f 0 (x). Plot f and f 0 .
0.4
Solution
Now f 0 (x) = cos x +
y
0.2
1
. Here is a plot.
x
0
Stewart 269/47
−0.2
6
function f(x) = sin(x) + ln(x)
derivative f’(x) = cos(x) + 1/x
5
−0.4
2
2.5
3
x
4
3.5
4
y
3
%-------------------------------------------------% Stewart 269/49
%
e = exp(1);
x = linspace(2, 4);
f = log(log(x));
TL = x ./ e - 1;
plot(x,f, x,TL,’r--’); grid on; hold on
plot(e, 0, ’go’, ’MarkerFaceColor’, ’g’, ...
’MarkerSize’, 7)
legend(’function f(x) = ln(ln(x))’, ...
’tangent line y = x/e - 1’, ’Location’, ’NorthWest’)
xlabel(’x’); ylabel(’y’)
title(’Stewart 269/49’)
%
2
1
0
−1
−2
0
2
4
6
x
8
10
12
%-------------------------------------------------% Stewart 269/47
%
syms x
f = sin(x) + log(x);
fp = diff(f,x); pretty(fp)
echo off; diary off
s269x56
cos(x) + 1/x
x = linspace(0.2, 12);
f = eval(vectorize(f));
fp = eval(vectorize(fp));
plot(x,f, x,fp,’r--’)
legend(’function f(x) = sin(x) + ln(x)’, ...
’derivative f’’(x) = cos(x) + 1/x’)
grid on; hold on
plot([0 12], [0 0], ’k’, ’LineWidth’, 1)
xlabel(’x’); ylabel(’y’)
title(’Stewart 269/47’)
%
Let y = x 8 ln x. Find
d9 y
, the ninth derivative of y.
dx9
Solution
An optional parameter to diff renders the needful in one line!
echo off; diary off
%-------------------------------------------------% Stewart 269/56
%
syms x
y = xˆ8 * log(x);
yp = diff(y,x,9); pretty(yp)
s269x49
Find an equation of the tangent line to the curve
y = f (x) = ln (ln x) at (x, y) = (e, 0).
40320
----x
Solution
%
echo off; diary off
• The slope of the tangent line is
1 1 1
=
m = f (e) =
ln x x x=e
e
0
• Therefore, an equation of the tangent line is
y−0
=
y
=
1
(x − e)
e
1 x − 1.
e
4