Consider the following logarithmic function.
y = ln[f(x)]
In a logarithm, if we have the spelling "ln", its a natural logarithm and its base is 'e'.
So, ln[f(x)] is a natural logarithm and its base is e.
Then, we have
y = lne[f(x)]
In lne[f(x)],
base = e
argument = f(x)
Wroking rule to find the derivative of a function which contains natural logarithm in the form y = ln[f(x)].
In y = ln[f(x)], since y is a function of x, we can find the derivative of y with respect to x, that is ᵈʸ⁄dₓ.
Step 1 :
y = ln[f(x)]
Let t = f(x).
Then, we have
y = ln(t)
Step 2 :
Now, y = ln(t) and t = f(x).
y = ln(t) ----> y is a function of t
t = f(x) ----> t is a function of x
Step 3 :
In step 2,
(i) since y is a function of t, we can find the derivative of y with respect to t.
(i) since t is a function of x, we can find the derivative of t with respect to x.
Chain rule to find derivative of y with respect to x.
Step 4 :
Substitute y = ln(t) and t = f(x).
To get the derivative of ln(t) with respect to t, write 1 in numerator and take the argument u in denominator. And the derivative f(x) with respect to x is f'(x).
Substitute t = f(x).
Working rule to find the derivative of a logarithmic function where the logarithm is not a natural logarithm :
Consider the following equation.
y = loga[f(x)]
Step 1 :
Convert the above equation to exponential form.
ay = f(x)
Step 2 :
Take natural logarithm on both sides.
ln(ay) = ln[f(x)]
Step 3 :
Use the power rule of logarithm.
yln(a) = ln[f(x)]
Step 4 :
Find the derivative on both sides.
Find ᵈʸ⁄dₓ in each of the following.
Problem 1 :
y = ln(x)
Solution :
y = ln(x)
Let t = x.
y = ln(t)
y = ln(t) ----> y is a function of t
t = x ----> t is a function of x
Chain Rule :
Substitute y = ln(t) and t = x.
Substitute t = x.
Problem 2 :
y = ln(2x)
Solution :
y = ln(2x)
Let t = 2x.
y = ln(t)
y = ln(t) ----> y is a function of t
t = 2x ----> t is a function of x
Chain Rule :
Substitute y = ln(t) and t = 2x.
Substitute t = 2x.
Problem 3 :
y = ln(x2 + 5x - 6)
Solution :
y = ln(x2 + 5x - 6)
Let t = x2 + 5x - 6.
y = ln(t)
y = ln(t) ----> y is a function of t
t = x2 + 5x - 6 ----> t is a function of x
Chain Rule :
Substitute y = ln(t) and t = x2 + 5x - 6.
Substitute t = x2 + 5x - 6.
Problem 4 :
y = ln(ex)
Solution :
y = ln(ex)
Let t = ex.
y = ln(t)
y = ln(t) ----> y is a function of t
t = ex ----> t is a function of x
Chain Rule :
Substitute y = ln(t) and t = ex.
Substitute t = ex.
Problem 5 :
y = ln(√x)
Solution :
y = ln(√x)
Let t = √x.
y = ln(t)
y = ln(t) ----> y is a function of t
t = √x ----> t is a function of x
Chain Rule :
Substitute y = ln(t) and t = √x.
Substitute t = √x.
Problem 6 :
y = ln[sin(x)]
Solution :
y = ln[sin(x)]
Let t = sin(x).
y = ln(t)
y = sin(x) ----> y is a function of t
t = √x ----> t is a function of x
Chain Rule :
Substitute y = ln(t) and t = sin(x).
Substitute t = sin(x).
Problem 7 :
y = ln[cos(√x)]
Solution :
y = ln[cos(√x)]
Let u = √x.
y = ln[cos(u)]
Let v = cos(u).
y = ln(v)
y = ln(v) ----> y is a function of v
v = cos(u) ----> v is a function of u
u = √x ----> u is a function of x
Chain Rule :
Substitute y = ln(v), v = cos(u) and u = √x.
Substitute v = cos(u).
Substitute u = √x.
Problem 8 :
Solution :
Let u = tan(x).
y = ln(√u)
Let v = √u.
y = ln(v)
y = ln(v) ----> y is a function of v
v = √u ----> v is a function of u
u = tan(x) ----> u is a function of x
Chain Rule :
Substitute y = ln(v), v = √u and u = tan(x).
Substitute v = √u.
Substitute u = tan(x).
Problem 9 :
y = log10(x)
Solution :
y = log10(x)
Convert the above equation to exponential form.
10y = x
Take natural logarithm on both sides.
ln(10y) = ln(x)
Use the power ruole of logarithm.
yln(10) = ln(x)
Find the derivative on both sides with respect to x.
Problem 10 :
y = log6(x2 + 3x + 5)
Solution :
y = log10(x2 + 3x + 5)
Convert the above equation to exponential form.
6y = x2 + 3x + 5
Take natural logarithm on both sides.
ln(6y) = ln(x2 + 3x + 5)
Use the power ruole of logarithm.
yln(6) = ln(x2 + 3x + 5)
Find the derivative on both sides with respect to x.
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