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Solutions to find the derivatives of various functions using logarithmic differentiation. The functions include logarithms, exponentials, and trigonometric functions. the concept of relative rate of change and demonstrates its application through examples. useful for students studying advanced mathematics, particularly calculus.
Typology: Lecture notes
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Recall that the logarithm functions satisfy very important arithmetic laws:
If a and b are positive numbers, and c is a positive number not equal to 1, we have:
logc a = ln a ln c
ln(ab) = ln a + ln b logc (ab) = logc a + logc b
ln
a b
= ln a − ln b logc
a b
= logc a − logc b
ln
ab
= b ln a logc
ab
= b logc a
The natural logarithm function was defined to the inverse of the exponential function e x^ , and the base a logarithms were defined to be the inverse of the exponential function a x^ , so we have the Cancellation Laws
eln^ x^ = x and aln^ x^ = x, so if we let y = ln x or y = loga x we have e y^ = x or a y^ = x.
Using the method of implicit differentiation, we get:
(e y^ )′^ = e y^ y′^ = (x)′^ = 1 or (ay^ )′^ = (ln a)ay^ y′^ = (x)′^ = 1,
so y′^ =
e y^
x or y′^ =
(ln a)ay^
(ln a)x
Thus
and
We may get more general formulas by using the Chain Rule:
and
The quantity f ′(x) f (x) is called the relative rate of change of the function f , and is
very important in practical applications.
y′^ = 1 x^5 + x^3 + 1 (x^5 + x^3 + 1 )′^ = 1 x^5 + x^3 + 1 ( 5 x^4 + 3 x^2 ) = x
(^2) ( 5 x (^2) + 3 ) x^5 + x^3 + 1
dx ( sin(ln(x)).
d dx (sin(ln(x)) = cos(ln x)
x
cos(ln x) x
d dx (ln (sin(ln(x))).
d dx (ln (sin(ln(x))) =
1 sin (ln(x)) (sin (ln(x)))′^ =
sin(ln(x)) cos(ln(x))(ln x)′^ =
sin(ln(x)) cos(ln(x))
x
cos(ln(x)) x sin(ln(x)
Note that the domain of the derivative function is much larger than that of the original function!
dx log 16 (sin x + cos x).
log 16 (sin x + cos x) = ln(sin x + cos x) ln 16 , so
d dx log 16 (sin x + cos x) =
ln 16
d dx ln(sin x + cos x) =
y′ y
x
x − 1
x + 2
which we need only simplify slightly to get y′^ in a usable form:
y′^ = y
x
x + 2
Since, by definition, x x^ =
eln^ x^
)x = e x^ ln^ x^ , the function is only defined for x > 0, and must always be positive.
We have ln y = ln (x x^ ) = x ln x, so
y′ y = (x ln x)′^ = (x)′(ln x) + x(ln x)′^ = ( 1 ) ln x + x
x = 1 + ln x,
so y′^ = x x^ ( 1 + ln x).
The sign of y′^ depends only on that of 1 + ln x, which is 0 if ln x = −1 or x = e−^1 = 1 e
If 0 < x <
e we have y′^ < 0 and if
e < x we have y′^ > 0.
There will be occasions when we wish to apply logarithms and deal with negative values of the variables concerned.
Of course, ln x is undefined if x ≤ 0. However, ln |x| is defined if x < 0.
Let us then find the derivative of ln |x| for non-zero x.
If x > 0, it is of course
x
If x < 0, then |x| = −x, so ln |x| = ln(−x), and we can apply the Chain Rule:
d dx (ln(−x)) =
−x
d dx (−x) =
−x
x
so we have the important formula