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MAT01A1: Derivatives of Log Functions andLogarithmic Differentiation
Dr Craig
Week: 4 May 2020
Reminder: the Chain Rule
If f and g are both differentiable and
F = f ◦ g is the composite function
defined by
F (x) = f (g(x)),
then F is differentiable and F ′ is given by
F ′(x) = f ′(g(x)).g′(x)
Two other reminders
I ddx (a
x) = ax ln a
I When we use implicit differentiation, we
regard y as a function of x.
In the slides that follow, I will sometimes use
y = lnx and sometimes y = `n x to denote
y = loge x. It is often good to use `n x when
writing by hand so you don’t confuse the
function with other similar symbols.
Derivatives of Log Functions
We can use implicit differentiation to find
the derivative of the log function y = loga x.
Proof: Let y = loga x.
∴ ay = x
d
dx(ay) =
d
dx(x)
∴ ay · ln a · y′ = 1
∴ y′ =1
ay · ln a
∴ y′ =1
x ln a
Proof: Let y = loga x.
∴ ay = x
d
dx(ay) =
d
dx(x)
∴ ay · ln a · y′ = 1
∴ y′ =1
ay · ln a
∴ y′ =1
x ln a
Proof: Let y = loga x.
∴ ay = x
d
dx(ay) =
d
dx(x)
∴ ay · ln a · y′ = 1
∴ y′ =1
ay · ln a
∴ y′ =1
x ln a
Proof: Let y = loga x.
∴ ay = x
d
dx(ay) =
d
dx(x)
∴ ay · ln a · y′ = 1
∴ y′ =1
ay · ln a
∴ y′ =1
x ln a
Proof: Let y = loga x.
∴ ay = x
d
dx(ay) =
d
dx(x)
∴ ay · ln a · y′ = 1
∴ y′ =1
ay · ln a
∴ y′ =1
x ln a
Proof: Let y = loga x.
∴ ay = x
d
dx(ay) =
d
dx(x)
∴ ay · ln a · y′ = 1
∴ y′ =1
ay · ln a
∴ y′ =1
x ln a
Derivatives of Log Functions
d
dx(loga x) =
1
x ln a
Notice that if a = e then we have
d
dx(lnx) =
1
x
An example (with chain rule):
Differentiate: y = log10(2 + cosx)
Derivatives of Log Functions
d
dx(loga x) =
1
x ln a
Notice that if a = e then we have
d
dx(lnx) =
1
x
An example (with chain rule):
Differentiate: y = log10(2 + cosx)
Derivatives of Log Functions
d
dx(loga x) =
1
x ln a
Notice that if a = e then we have
d
dx(lnx) =
1
x
An example (with chain rule):
Differentiate: y = log10(2 + cosx)
Example: Differentiate y = log10(2 + cosx).
Solution:
d
dxy =
d
dx(log10(2 + cosx))
∴ y′ =1
(2 + cosx) ln 10· ddx
(2 + cosx)
∴ y′ =− sinx
(2 + cosx) ln 10
Example: Differentiate y = log10(2 + cosx).
Solution:
d
dxy =
d
dx(log10(2 + cosx))
∴ y′ =1
(2 + cosx) ln 10· ddx
(2 + cosx)
∴ y′ =− sinx
(2 + cosx) ln 10
Example: Differentiate y = log10(2 + cosx).
Solution:
d
dxy =
d
dx(log10(2 + cosx))
∴ y′ =1
(2 + cosx) ln 10· ddx
(2 + cosx)
∴ y′ =− sinx
(2 + cosx) ln 10
Example: Differentiate y = log10(2 + cosx).
Solution:
d
dxy =
d
dx(log10(2 + cosx))
∴ y′ =1
(2 + cosx) ln 10· ddx
(2 + cosx)
∴ y′ =− sinx
(2 + cosx) ln 10
Derivative of y = lnx
d
dx(loga x) =
1
x ln a⇒ d
dx(lnx) =
1
x
Example: Differentiate y = ln(x3 + 1).
d
dx(ln(x3+1)) =
1
x3 + 1· ddx
(x3+1) =3x2
x3 + 1
In general, if we combine derivatives of log
functions with the chain rule, we get
d
dx(ln(g(x))) =
g′(x)
g(x)
Derivative of y = lnx
d
dx(loga x) =
1
x ln a⇒ d
dx(lnx) =
1
x
Example: Differentiate y = ln(x3 + 1).
d
dx(ln(x3+1)) =
1
x3 + 1· ddx
(x3+1) =3x2
x3 + 1
In general, if we combine derivatives of log
functions with the chain rule, we get
d
dx(ln(g(x))) =
g′(x)
g(x)
Derivative of y = lnx
d
dx(loga x) =
1
x ln a⇒ d
dx(lnx) =
1
x
Example: Differentiate y = ln(x3 + 1).
d
dx(ln(x3+1)) =
1
x3 + 1· ddx
(x3+1) =3x2
x3 + 1
In general, if we combine derivatives of log
functions with the chain rule, we get
d
dx(ln(g(x))) =
g′(x)
g(x)
Derivative of y = lnx
d
dx(loga x) =
1
x ln a⇒ d
dx(lnx) =
1
x
Example: Differentiate y = ln(x3 + 1).
d
dx(ln(x3+1)) =
1
x3 + 1· ddx
(x3+1) =3x2
x3 + 1
In general, if we combine derivatives of log
functions with the chain rule, we get
d
dx(ln(g(x))) =
g′(x)
g(x)
More examples:
Differentiate the following:
1. y = ln(sinx)
2. f (x) =√lnx
3. g(x) = ln
(x + 1√x− 2
)
Solutions:
1.d
dx(ln(sinx)) =
1
sinx· ddx
(sinx) = cotx
More examples:
Differentiate the following:
1. y = ln(sinx)
2. f (x) =√lnx
3. g(x) = ln
(x + 1√x− 2
)Solutions:
1.d
dx(ln(sinx)) =
1
sinx· ddx
(sinx) = cotx
More examples:
Differentiate the following:
1. y = ln(sinx)
2. f (x) =√lnx
3. g(x) = ln
(x + 1√x− 2
)Solutions:
1.d
dx(ln(sinx)) =
1
sinx· ddx
(sinx) = cotx
Solutions:
2. f ′(x) =1
2√lnx· ddx
(lnx) =1
2x√lnx
3.d
dx
(ln
x + 1√x− 2
)=
d
dx
(ln(x + 1)− ln
√x− 2
)=
d
dxln(x + 1)− 1
2
d
dxln(x− 2)
=1
x + 1− 1
2(x− 2)=
x− 5
2(x + 1)(x− 2)
Solutions:
2. f ′(x) =1
2√lnx· ddx
(lnx) =1
2x√lnx
3.d
dx
(ln
x + 1√x− 2
)
=d
dx
(ln(x + 1)− ln
√x− 2
)=
d
dxln(x + 1)− 1
2
d
dxln(x− 2)
=1
x + 1− 1
2(x− 2)=
x− 5
2(x + 1)(x− 2)
Solutions:
2. f ′(x) =1
2√lnx· ddx
(lnx) =1
2x√lnx
3.d
dx
(ln
x + 1√x− 2
)=
d
dx
(ln(x + 1)− ln
√x− 2
)
=d
dxln(x + 1)− 1
2
d
dxln(x− 2)
=1
x + 1− 1
2(x− 2)=
x− 5
2(x + 1)(x− 2)
Solutions:
2. f ′(x) =1
2√lnx· ddx
(lnx) =1
2x√lnx
3.d
dx
(ln
x + 1√x− 2
)=
d
dx
(ln(x + 1)− ln
√x− 2
)=
d
dxln(x + 1)− 1
2
d
dxln(x− 2)
=1
x + 1− 1
2(x− 2)=
x− 5
2(x + 1)(x− 2)
Solutions:
2. f ′(x) =1
2√lnx· ddx
(lnx) =1
2x√lnx
3.d
dx
(ln
x + 1√x− 2
)=
d
dx
(ln(x + 1)− ln
√x− 2
)=
d
dxln(x + 1)− 1
2
d
dxln(x− 2)
=1
x + 1− 1
2(x− 2)
=x− 5
2(x + 1)(x− 2)
Solutions:
2. f ′(x) =1
2√lnx· ddx
(lnx) =1
2x√lnx
3.d
dx
(ln
x + 1√x− 2
)=
d
dx
(ln(x + 1)− ln
√x− 2
)=
d
dxln(x + 1)− 1
2
d
dxln(x− 2)
=1
x + 1− 1
2(x− 2)=
x− 5
2(x + 1)(x− 2)
A graph of the last example:
d
dx
(ln
x+ 1√x− 2
)=
x− 5
2(x+ 1)(x− 2)(x > 2)
Differentiation so far:
I Product rule: (f.g)′ = f ′g + g′f
I Quotient rule:
(f
g
)′=
f ′g − g′f
g2
I Trig derivatives (from special limits)
I Chain Rule: F = f ◦ g,
F ′(x) = f ′(g(x)).g′(x)
I Implicit differentiation: treat y as a
function of x
I Inverse Trig derivatives
I Derivatives of Log functions
The function f (x) = `n(x) is only defined
for x > 0. Hence its derivative f ′(x) =1
xis
only defined for x > 0. What about the
function f (x) = `n|x|?
f (x) = `n|x| f ′(x) =1
x(both defined for x ∈ (−∞, 0) ∪ (0,∞))
What do we know so far about ddx(a
b)?
Let y = ab.
I If a, b ∈ R, then y′ = 0.
I If a = f (x), b ∈ R, then we apply the
chain rule with the power rule to get
y′ =d
dx
(f (x)b
)= b[f (x)]b−1f ′(x)
I If a ∈ R, b = g(x), we apply the chain
rule and ddx(a
x) to get
y′ =d
dx
(ag(x)
)= ag(x)(ln a)g′(x)
What do we know so far about ddx(a
b)?
Let y = ab.
I If a, b ∈ R, then y′ = 0.
I If a = f (x), b ∈ R, then we apply the
chain rule with the power rule to get
y′ =d
dx
(f (x)b
)= b[f (x)]b−1f ′(x)
I If a ∈ R, b = g(x), we apply the chain
rule and ddx(a
x) to get
y′ =d
dx
(ag(x)
)= ag(x)(ln a)g′(x)
What do we know so far about ddx(a
b)?
Let y = ab.
I If a, b ∈ R, then y′ = 0.
I If a = f (x), b ∈ R, then we apply the
chain rule with the power rule to get
y′ =d
dx
(f (x)b
)= b[f (x)]b−1f ′(x)
I If a ∈ R, b = g(x), we apply the chain
rule and ddx(a
x) to get
y′ =d
dx
(ag(x)
)= ag(x)(ln a)g′(x)
What do we know so far about ddx(a
b)?
Let y = ab.
I If a, b ∈ R, then y′ = 0.
I If a = f (x), b ∈ R, then we apply the
chain rule with the power rule to get
y′ =d
dx
(f (x)b
)= b[f (x)]b−1f ′(x)
I If a ∈ R, b = g(x), we apply the chain
rule and ddx(a
x) to get
y′ =d
dx
(ag(x)
)= ag(x)(ln a)g′(x)
Logarithmic Differentiation:
Consider the function
y = x2x+3
How do we differentiate this? Do we use the
power rule? Or the rule for exponential
functions?
Neither!
Logarithmic Differentiation:
Consider the function
y = x2x+3
How do we differentiate this?
Do we use the
power rule? Or the rule for exponential
functions?
Neither!
Logarithmic Differentiation:
Consider the function
y = x2x+3
How do we differentiate this? Do we use the
power rule?
Or the rule for exponential
functions?
Neither!
Logarithmic Differentiation:
Consider the function
y = x2x+3
How do we differentiate this? Do we use the
power rule? Or the rule for exponential
functions?
Neither!
Logarithmic Differentiation:
Consider the function
y = x2x+3
How do we differentiate this? Do we use the
power rule? Or the rule for exponential
functions?
Neither!
Steps in logarithmic differentiation:
1. Take natural logarithms of both sides of
y = f (x) and use the Log Laws to
simplify the result.
2. Differentiate implicitly with respect to x.
3. Solve for y′.
Steps in logarithmic differentiation:
1. Take natural logarithms of both sides of
y = f (x) and use the Log Laws to
simplify the result.
2. Differentiate implicitly with respect to x.
3. Solve for y′.
Steps in logarithmic differentiation:
1. Take natural logarithms of both sides of
y = f (x) and use the Log Laws to
simplify the result.
2. Differentiate implicitly with respect to x.
3. Solve for y′.
Example: Differentiate y = (cosx)x
∴ `n(y) = `n ((cosx)x)
= x · `n(cosx)
Now we differentiate both sides:
∴d
dx(`n(y)) =
d
dx(x)`n(cosx) + x
d
dx(`n(cosx))
∴y′
y= `n(cosx) + x · − sinx
cosx
∴ y′ = y (`n(cosx)− x · tanx)∴ y′ = (cosx)x (`n(cosx)− x · tanx)
Example: Differentiate y = (cosx)x
∴ `n(y) = `n ((cosx)x)
= x · `n(cosx)
Now we differentiate both sides:
∴d
dx(`n(y)) =
d
dx(x)`n(cosx) + x
d
dx(`n(cosx))
∴y′
y= `n(cosx) + x · − sinx
cosx
∴ y′ = y (`n(cosx)− x · tanx)∴ y′ = (cosx)x (`n(cosx)− x · tanx)
Example: Differentiate y = (cosx)x
∴ `n(y) = `n ((cosx)x)
= x · `n(cosx)
Now we differentiate both sides:
∴d
dx(`n(y)) =
d
dx(x)`n(cosx) + x
d
dx(`n(cosx))
∴y′
y= `n(cosx) + x · − sinx
cosx
∴ y′ = y (`n(cosx)− x · tanx)∴ y′ = (cosx)x (`n(cosx)− x · tanx)
Example: Differentiate y = (cosx)x
∴ `n(y) = `n ((cosx)x)
= x · `n(cosx)
Now we differentiate both sides:
∴d
dx(`n(y)) =
d
dx(x)`n(cosx) + x
d
dx(`n(cosx))
∴y′
y= `n(cosx) + x · − sinx
cosx
∴ y′ = y (`n(cosx)− x · tanx)∴ y′ = (cosx)x (`n(cosx)− x · tanx)
Example: Differentiate y = (cosx)x
∴ `n(y) = `n ((cosx)x)
= x · `n(cosx)
Now we differentiate both sides:
∴d
dx(`n(y)) =
d
dx(x)`n(cosx) + x
d
dx(`n(cosx))
∴y′
y= `n(cosx) + x · − sinx
cosx
∴ y′ = y (`n(cosx)− x · tanx)∴ y′ = (cosx)x (`n(cosx)− x · tanx)
Example: Differentiate y = (cosx)x
∴ `n(y) = `n ((cosx)x)
= x · `n(cosx)
Now we differentiate both sides:
∴d
dx(`n(y)) =
d
dx(x)`n(cosx) + x
d
dx(`n(cosx))
∴y′
y= `n(cosx) + x · − sinx
cosx
∴ y′ = y (`n(cosx)− x · tanx)∴ y′ = (cosx)x (`n(cosx)− x · tanx)
Example: Differentiate y = (cosx)x
∴ `n(y) = `n ((cosx)x)
= x · `n(cosx)
Now we differentiate both sides:
∴d
dx(`n(y)) =
d
dx(x)`n(cosx) + x
d
dx(`n(cosx))
∴y′
y= `n(cosx) + x · − sinx
cosx
∴ y′ = y (`n(cosx)− x · tanx)
∴ y′ = (cosx)x (`n(cosx)− x · tanx)
Example: Differentiate y = (cosx)x
∴ `n(y) = `n ((cosx)x)
= x · `n(cosx)
Now we differentiate both sides:
∴d
dx(`n(y)) =
d
dx(x)`n(cosx) + x
d
dx(`n(cosx))
∴y′
y= `n(cosx) + x · − sinx
cosx
∴ y′ = y (`n(cosx)− x · tanx)∴ y′ = (cosx)x (`n(cosx)− x · tanx)
More examples:
y = (sinx)x3
and y = x√x
Try to find y′ on your own first. The
solutions are on the next two slides.
Example: Differentiate y = (sinx)x3.
Solution:
ln(y)=ln((sinx)x
3)=x3 ln(sinx)
Differentiating both sides gives:
d
dx(ln(y)) =
d
dx(x3) ln(sinx) + x3 · d
dx(ln(sinx))
∴y′
y= 3x2 ln(sinx) + x3
1
sinx· ddx
(sinx)
∴ y′ = y(3x2 ln(sinx) + x3 cotx
)∴ y′ = (sinx)x
3(3x2 ln(sinx) + x3 cotx
)
Example: Differentiate y = (sinx)x3.
Solution: ln(y)=ln((sinx)x
3)=x3 ln(sinx)
Differentiating both sides gives:
d
dx(ln(y)) =
d
dx(x3) ln(sinx) + x3 · d
dx(ln(sinx))
∴y′
y= 3x2 ln(sinx) + x3
1
sinx· ddx
(sinx)
∴ y′ = y(3x2 ln(sinx) + x3 cotx
)∴ y′ = (sinx)x
3(3x2 ln(sinx) + x3 cotx
)
Example: Differentiate y = (sinx)x3.
Solution: ln(y)=ln((sinx)x
3)=x3 ln(sinx)
Differentiating both sides gives:
d
dx(ln(y)) =
d
dx(x3) ln(sinx) + x3 · d
dx(ln(sinx))
∴y′
y= 3x2 ln(sinx) + x3
1
sinx· ddx
(sinx)
∴ y′ = y(3x2 ln(sinx) + x3 cotx
)∴ y′ = (sinx)x
3(3x2 ln(sinx) + x3 cotx
)
Example: Differentiate y = (sinx)x3.
Solution: ln(y)=ln((sinx)x
3)=x3 ln(sinx)
Differentiating both sides gives:
d
dx(ln(y)) =
d
dx(x3) ln(sinx) + x3 · d
dx(ln(sinx))
∴y′
y= 3x2 ln(sinx) + x3
1
sinx· ddx
(sinx)
∴ y′ = y(3x2 ln(sinx) + x3 cotx
)∴ y′ = (sinx)x
3(3x2 ln(sinx) + x3 cotx
)
Example: Differentiate y = (sinx)x3.
Solution: ln(y)=ln((sinx)x
3)=x3 ln(sinx)
Differentiating both sides gives:
d
dx(ln(y)) =
d
dx(x3) ln(sinx) + x3 · d
dx(ln(sinx))
∴y′
y= 3x2 ln(sinx) + x3
1
sinx· ddx
(sinx)
∴ y′ = y(3x2 ln(sinx) + x3 cotx
)∴ y′ = (sinx)x
3(3x2 ln(sinx) + x3 cotx
)
Example: Differentiate y = (sinx)x3.
Solution: ln(y)=ln((sinx)x
3)=x3 ln(sinx)
Differentiating both sides gives:
d
dx(ln(y)) =
d
dx(x3) ln(sinx) + x3 · d
dx(ln(sinx))
∴y′
y= 3x2 ln(sinx) + x3
1
sinx· ddx
(sinx)
∴ y′ = y(3x2 ln(sinx) + x3 cotx
)
∴ y′ = (sinx)x3(3x2 ln(sinx) + x3 cotx
)
Example: Differentiate y = (sinx)x3.
Solution: ln(y)=ln((sinx)x
3)=x3 ln(sinx)
Differentiating both sides gives:
d
dx(ln(y)) =
d
dx(x3) ln(sinx) + x3 · d
dx(ln(sinx))
∴y′
y= 3x2 ln(sinx) + x3
1
sinx· ddx
(sinx)
∴ y′ = y(3x2 ln(sinx) + x3 cotx
)∴ y′ = (sinx)x
3(3x2 ln(sinx) + x3 cotx
)
Example: Differentiate y = x√x.
Solution:
`n(y) = `n(x√x)=√x `n(x)
Differentiating both sides gives:
d
dx(`n(y)) =
d
dx(√x)`n(x) +
√x · d
dx(`n(x))
∴y′
y=
1
2√x`n(x) +
√x · 1
x
∴y′
y=
`n(x)
2√x
+1√x
∴ y′ =(x√x)(`n(x) + 2√
x
)
Example: Differentiate y = x√x.
Solution: `n(y) = `n(x√x)=√x `n(x)
Differentiating both sides gives:
d
dx(`n(y)) =
d
dx(√x)`n(x) +
√x · d
dx(`n(x))
∴y′
y=
1
2√x`n(x) +
√x · 1
x
∴y′
y=
`n(x)
2√x
+1√x
∴ y′ =(x√x)(`n(x) + 2√
x
)
Example: Differentiate y = x√x.
Solution: `n(y) = `n(x√x)=√x `n(x)
Differentiating both sides gives:
d
dx(`n(y)) =
d
dx(√x)`n(x) +
√x · d
dx(`n(x))
∴y′
y=
1
2√x`n(x) +
√x · 1
x
∴y′
y=
`n(x)
2√x
+1√x
∴ y′ =(x√x)(`n(x) + 2√
x
)
Example: Differentiate y = x√x.
Solution: `n(y) = `n(x√x)=√x `n(x)
Differentiating both sides gives:
d
dx(`n(y)) =
d
dx(√x)`n(x) +
√x · d
dx(`n(x))
∴y′
y=
1
2√x`n(x) +
√x · 1
x
∴y′
y=
`n(x)
2√x
+1√x
∴ y′ =(x√x)(`n(x) + 2√
x
)
Example: Differentiate y = x√x.
Solution: `n(y) = `n(x√x)=√x `n(x)
Differentiating both sides gives:
d
dx(`n(y)) =
d
dx(√x)`n(x) +
√x · d
dx(`n(x))
∴y′
y=
1
2√x`n(x) +
√x · 1
x
∴y′
y=
`n(x)
2√x
+1√x
∴ y′ =(x√x)(`n(x) + 2√
x
)
Example: Differentiate y = x√x.
Solution: `n(y) = `n(x√x)=√x `n(x)
Differentiating both sides gives:
d
dx(`n(y)) =
d
dx(√x)`n(x) +
√x · d
dx(`n(x))
∴y′
y=
1
2√x`n(x) +
√x · 1
x
∴y′
y=
`n(x)
2√x
+1√x
∴ y′ =(x√x)(`n(x) + 2√
x
)
Example: Differentiate y = x√x.
Solution: `n(y) = `n(x√x)=√x `n(x)
Differentiating both sides gives:
d
dx(`n(y)) =
d
dx(√x)`n(x) +
√x · d
dx(`n(x))
∴y′
y=
1
2√x`n(x) +
√x · 1
x
∴y′
y=
`n(x)
2√x
+1√x
∴ y′ =(x√x)(`n(x) + 2√
x
)
Another useful application:
Suppose we have a function like:
y =x3/4√x2 + 1
(3x + 2)5
We could differentiate this thinking of
y = f(x)g(x) , applying the quotient rule, and
then applying the product rule to find f ′(x).
This method will work but it will be very
tough and we are likely to make a mistake
somewhere along the way. So, instead, we
use logarithmic differentiation.
Another useful application:
Suppose we have a function like:
y =x3/4√x2 + 1
(3x + 2)5
We could differentiate this thinking of
y = f(x)g(x) , applying the quotient rule, and
then applying the product rule to find f ′(x).
This method will work but it will be very
tough and we are likely to make a mistake
somewhere along the way. So, instead, we
use logarithmic differentiation.
Another useful application:
Suppose we have a function like:
y =x3/4√x2 + 1
(3x + 2)5
We could differentiate this thinking of
y = f(x)g(x) , applying the quotient rule, and
then applying the product rule to find f ′(x).
This method will work but it will be very
tough and we are likely to make a mistake
somewhere along the way.
So, instead, we
use logarithmic differentiation.
Another useful application:
Suppose we have a function like:
y =x3/4√x2 + 1
(3x + 2)5
We could differentiate this thinking of
y = f(x)g(x) , applying the quotient rule, and
then applying the product rule to find f ′(x).
This method will work but it will be very
tough and we are likely to make a mistake
somewhere along the way. So, instead, we
use logarithmic differentiation.
Logarithmic differentiation
y =x3/4√x2 + 1
(3x + 2)5
1. Take the natural logarithm of both sides
and simplify with Log Laws.
2. Differentiate both sides with respect to x
(implicitly on the LHS).
3. Solve for y′.
It is important to remember that
d
dx
(ln(g(x))
)=
g′(x)
g(x)
Logarithmic differentiation
y =x3/4√x2 + 1
(3x + 2)5
1. Take the natural logarithm of both sides
and simplify with Log Laws.
2. Differentiate both sides with respect to x
(implicitly on the LHS).
3. Solve for y′.
It is important to remember that
d
dx
(ln(g(x))
)=
g′(x)
g(x)
y =x3/4√x2 + 1
(3x + 2)5
∴ `n(y) = `n
(x3/4√x2 + 1
(3x + 2)5
)= `n
(x3/4
)+ `n
(√x2 + 1
)− `n
((3x + 2)5
)=
3
4`n(x) +
1
2`n(x2 + 1)− 5`n(3x + 2)
y =x3/4√x2 + 1
(3x + 2)5
∴ `n(y) = `n
(x3/4√x2 + 1
(3x + 2)5
)
= `n(x3/4
)+ `n
(√x2 + 1
)− `n
((3x + 2)5
)=
3
4`n(x) +
1
2`n(x2 + 1)− 5`n(3x + 2)
y =x3/4√x2 + 1
(3x + 2)5
∴ `n(y) = `n
(x3/4√x2 + 1
(3x + 2)5
)= `n
(x3/4
)+ `n
(√x2 + 1
)− `n
((3x + 2)5
)
=3
4`n(x) +
1
2`n(x2 + 1)− 5`n(3x + 2)
y =x3/4√x2 + 1
(3x + 2)5
∴ `n(y) = `n
(x3/4√x2 + 1
(3x + 2)5
)= `n
(x3/4
)+ `n
(√x2 + 1
)− `n
((3x + 2)5
)=
3
4`n(x) +
1
2`n(x2 + 1)− 5`n(3x + 2)
We have
`n(y) =3
4`n(x)+
1
2`n(x2+1)− 5`n(3x+2)
Now we differentiate both sides w.r.t x
∴y′
y=
3
4x+1
2
(1
x2 + 1
)(2x)−5
(1
3x + 2
)(3)
∴y′
y=
3
4x+
x
x2 + 1− 15
3x + 2
∴ y′ =x3/4√x2 + 1
(3x + 2)5
(3
4x+
x
x2 + 1− 15
3x + 2
)
We have
`n(y) =3
4`n(x)+
1
2`n(x2+1)− 5`n(3x+2)
Now we differentiate both sides w.r.t x
∴y′
y=
3
4x+1
2
(1
x2 + 1
)(2x)−5
(1
3x + 2
)(3)
∴y′
y=
3
4x+
x
x2 + 1− 15
3x + 2
∴ y′ =x3/4√x2 + 1
(3x + 2)5
(3
4x+
x
x2 + 1− 15
3x + 2
)
We have
`n(y) =3
4`n(x)+
1
2`n(x2+1)− 5`n(3x+2)
Now we differentiate both sides w.r.t x
∴y′
y=
3
4x+1
2
(1
x2 + 1
)(2x)−5
(1
3x + 2
)(3)
∴y′
y=
3
4x+
x
x2 + 1− 15
3x + 2
∴ y′ =x3/4√x2 + 1
(3x + 2)5
(3
4x+
x
x2 + 1− 15
3x + 2
)
We have
`n(y) =3
4`n(x)+
1
2`n(x2+1)− 5`n(3x+2)
Now we differentiate both sides w.r.t x
∴y′
y=
3
4x+1
2
(1
x2 + 1
)(2x)−5
(1
3x + 2
)(3)
∴y′
y=
3
4x+
x
x2 + 1− 15
3x + 2
∴ y′ =x3/4√x2 + 1
(3x + 2)5
(3
4x+
x
x2 + 1− 15
3x + 2
)
We have
`n(y) =3
4`n(x)+
1
2`n(x2+1)− 5`n(3x+2)
Now we differentiate both sides w.r.t x
∴y′
y=
3
4x+1
2
(1
x2 + 1
)(2x)−5
(1
3x + 2
)(3)
∴y′
y=
3
4x+
x
x2 + 1− 15
3x + 2
∴ y′ =x3/4√x2 + 1
(3x + 2)5
(3
4x+
x
x2 + 1− 15
3x + 2
)
Logarithmic differentiation
Here are two more examples to which you
can apply logarithmic differentiation.
Attempt them on your own before looking at
the solutions:
1. f (x) =(x + 1)10
(2x− 4)8
2. h(x) =x8 cos3 x√
x− 1
Solution (1.):
`n(f (x)) = `n
((x + 1)10
(2x− 4)8
)∴ `n(f (x)) = `n
((x + 1)10
)− `n
((2x− 4)8
)∴ `n(f (x)) = 10`n(x + 1)− 8`n(2x− 4)
Now we differentiate both sides:
∴f ′(x)
f (x)=
10
x + 1− 16
2x− 4
∴ f ′(x) =(x + 1)10
(2x− 4)8
(10
x + 1− 8
x− 2
)
Solution (1.): `n(f (x)) = `n
((x + 1)10
(2x− 4)8
)
∴ `n(f (x)) = `n((x + 1)10
)− `n
((2x− 4)8
)∴ `n(f (x)) = 10`n(x + 1)− 8`n(2x− 4)
Now we differentiate both sides:
∴f ′(x)
f (x)=
10
x + 1− 16
2x− 4
∴ f ′(x) =(x + 1)10
(2x− 4)8
(10
x + 1− 8
x− 2
)
Solution (1.): `n(f (x)) = `n
((x + 1)10
(2x− 4)8
)∴ `n(f (x)) = `n
((x + 1)10
)− `n
((2x− 4)8
)
∴ `n(f (x)) = 10`n(x + 1)− 8`n(2x− 4)
Now we differentiate both sides:
∴f ′(x)
f (x)=
10
x + 1− 16
2x− 4
∴ f ′(x) =(x + 1)10
(2x− 4)8
(10
x + 1− 8
x− 2
)
Solution (1.): `n(f (x)) = `n
((x + 1)10
(2x− 4)8
)∴ `n(f (x)) = `n
((x + 1)10
)− `n
((2x− 4)8
)∴ `n(f (x)) = 10`n(x + 1)− 8`n(2x− 4)
Now we differentiate both sides:
∴f ′(x)
f (x)=
10
x + 1− 16
2x− 4
∴ f ′(x) =(x + 1)10
(2x− 4)8
(10
x + 1− 8
x− 2
)
Solution (1.): `n(f (x)) = `n
((x + 1)10
(2x− 4)8
)∴ `n(f (x)) = `n
((x + 1)10
)− `n
((2x− 4)8
)∴ `n(f (x)) = 10`n(x + 1)− 8`n(2x− 4)
Now we differentiate both sides:
∴f ′(x)
f (x)=
10
x + 1− 16
2x− 4
∴ f ′(x) =(x + 1)10
(2x− 4)8
(10
x + 1− 8
x− 2
)
Solution (1.): `n(f (x)) = `n
((x + 1)10
(2x− 4)8
)∴ `n(f (x)) = `n
((x + 1)10
)− `n
((2x− 4)8
)∴ `n(f (x)) = 10`n(x + 1)− 8`n(2x− 4)
Now we differentiate both sides:
∴f ′(x)
f (x)=
10
x + 1− 16
2x− 4
∴ f ′(x) =(x + 1)10
(2x− 4)8
(10
x + 1− 8
x− 2
)
Solution (1.): `n(f (x)) = `n
((x + 1)10
(2x− 4)8
)∴ `n(f (x)) = `n
((x + 1)10
)− `n
((2x− 4)8
)∴ `n(f (x)) = 10`n(x + 1)− 8`n(2x− 4)
Now we differentiate both sides:
∴f ′(x)
f (x)=
10
x + 1− 16
2x− 4
∴ f ′(x) =(x + 1)10
(2x− 4)8
(10
x + 1− 8
x− 2
)
The answer we have below is fine as a final
answer.
f ′(x) =(x + 1)10
(2x− 4)8
(10
x + 1− 8
x− 2
)If you were feeling energetic you could
simplify it further to
f ′(x) =(x + 1)9(x− 14)
128(x− 2)9
Solution (2.):
`n(h(x)) = `n(x8 cos3 x√
x−1
)∴ `n(h(x)) = `n(x8) + `n(cos3 x)− `n
(√x− 1
)∴ `n(h(x)) = 8`n(x)+3`n(cosx)−1
2`n(x−1)
Now we differentiate both sides:
∴h′(x)
h(x)=
8
x+ 3
(− sinx)
cosx− 1
2(x− 1)
∴ h′(x) =
(x8 cos3 x√
x− 1
)(8
x− 3 tanx− 1
2(x− 1)
)
Solution (2.): `n(h(x)) = `n(x8 cos3 x√
x−1
)
∴ `n(h(x)) = `n(x8) + `n(cos3 x)− `n(√
x− 1)
∴ `n(h(x)) = 8`n(x)+3`n(cosx)−12`n(x−1)
Now we differentiate both sides:
∴h′(x)
h(x)=
8
x+ 3
(− sinx)
cosx− 1
2(x− 1)
∴ h′(x) =
(x8 cos3 x√
x− 1
)(8
x− 3 tanx− 1
2(x− 1)
)
Solution (2.): `n(h(x)) = `n(x8 cos3 x√
x−1
)∴ `n(h(x)) = `n(x8) + `n(cos3 x)− `n
(√x− 1
)
∴ `n(h(x)) = 8`n(x)+3`n(cosx)−12`n(x−1)
Now we differentiate both sides:
∴h′(x)
h(x)=
8
x+ 3
(− sinx)
cosx− 1
2(x− 1)
∴ h′(x) =
(x8 cos3 x√
x− 1
)(8
x− 3 tanx− 1
2(x− 1)
)
Solution (2.): `n(h(x)) = `n(x8 cos3 x√
x−1
)∴ `n(h(x)) = `n(x8) + `n(cos3 x)− `n
(√x− 1
)∴ `n(h(x)) = 8`n(x)+3`n(cosx)−1
2`n(x−1)
Now we differentiate both sides:
∴h′(x)
h(x)=
8
x+ 3
(− sinx)
cosx− 1
2(x− 1)
∴ h′(x) =
(x8 cos3 x√
x− 1
)(8
x− 3 tanx− 1
2(x− 1)
)
Solution (2.): `n(h(x)) = `n(x8 cos3 x√
x−1
)∴ `n(h(x)) = `n(x8) + `n(cos3 x)− `n
(√x− 1
)∴ `n(h(x)) = 8`n(x)+3`n(cosx)−1
2`n(x−1)
Now we differentiate both sides:
∴h′(x)
h(x)=
8
x+ 3
(− sinx)
cosx− 1
2(x− 1)
∴ h′(x) =
(x8 cos3 x√
x− 1
)(8
x− 3 tanx− 1
2(x− 1)
)
Solution (2.): `n(h(x)) = `n(x8 cos3 x√
x−1
)∴ `n(h(x)) = `n(x8) + `n(cos3 x)− `n
(√x− 1
)∴ `n(h(x)) = 8`n(x)+3`n(cosx)−1
2`n(x−1)
Now we differentiate both sides:
∴h′(x)
h(x)=
8
x+ 3
(− sinx)
cosx− 1
2(x− 1)
∴ h′(x) =
(x8 cos3 x√
x− 1
)(8
x− 3 tanx− 1
2(x− 1)
)
Solution (2.): `n(h(x)) = `n(x8 cos3 x√
x−1
)∴ `n(h(x)) = `n(x8) + `n(cos3 x)− `n
(√x− 1
)∴ `n(h(x)) = 8`n(x)+3`n(cosx)−1
2`n(x−1)
Now we differentiate both sides:
∴h′(x)
h(x)=
8
x+ 3
(− sinx)
cosx− 1
2(x− 1)
∴ h′(x) =
(x8 cos3 x√
x− 1
)(8
x− 3 tanx− 1
2(x− 1)
)
These last few slides are here just to give you
a better understanding of the number e.
Previously, we introduced e as follows:
e is the number such that
limh→0
eh − 1
h= 1
In other words, it is the number such that the
curve y = ex has a gradient of 1 at x = 0.
Now we want to express e as a limit.
These last few slides are here just to give you
a better understanding of the number e.
Previously, we introduced e as follows:
e is the number such that
limh→0
eh − 1
h= 1
In other words, it is the number such that the
curve y = ex has a gradient of 1 at x = 0.
Now we want to express e as a limit.
These last few slides are here just to give you
a better understanding of the number e.
Previously, we introduced e as follows:
e is the number such that
limh→0
eh − 1
h= 1
In other words, it is the number such that the
curve y = ex has a gradient of 1 at x = 0.
Now we want to express e as a limit.
The number e as a limit
Let f (x) = `n(x). We know that f ′(x) =1
xand hence f ′(1) = 1/1 = 1.
Therefore
1 = f ′(1) = limh→0
f (1 + h)− f (1)
h
= limh→0
`n(1 + h)− `n(1)
h
= limx→0
`n(1 + x)− `n(1)
x
= limx→0
1
x`n(1 + x) = lim
x→0`n(1 + x)1/x
The number e as a limit
Let f (x) = `n(x). We know that f ′(x) =1
xand hence f ′(1) = 1/1 = 1. Therefore
1 = f ′(1) = limh→0
f (1 + h)− f (1)
h
= limh→0
`n(1 + h)− `n(1)
h
= limx→0
`n(1 + x)− `n(1)
x
= limx→0
1
x`n(1 + x) = lim
x→0`n(1 + x)1/x
Now,
e = e1 = elimx→0 `n(1+x)1/x
= limx→0
e`n(1+x)1/x
= limx→0
(1 + x)1/x
That is,
e = limx→0
(1 + x)1/x
As x→ 0, the bracket (1 + x) gets very close
to 1, but the 1/x will become very large.
Now,
e = e1 = elimx→0 `n(1+x)1/x
= limx→0
e`n(1+x)1/x
= limx→0
(1 + x)1/x
That is,
e = limx→0
(1 + x)1/x
As x→ 0, the bracket (1 + x) gets very close
to 1, but the 1/x will become very large.
Now,
e = e1 = elimx→0 `n(1+x)1/x
= limx→0
e`n(1+x)1/x
= limx→0
(1 + x)1/x
That is,
e = limx→0
(1 + x)1/x
As x→ 0, the bracket (1 + x) gets very close
to 1, but the 1/x will become very large.
We have that e = limx→0
(1 + x)1/x.
Here are some calculations for x→ 0+.
Prescribed tut problems:
Complete the following exercises from the
8th edition:
Ch 3.6:
2, 5, 8, 9, 13, 20, 29, 34, 39, 41, 44, 49, 52
If you are using the 7th edition:
Ch 3.6: 13 → 11