Logarithmic Differentiation (original) (raw)

Last Updated : 26 Nov, 2025

Logarithmic differentiation is a technique used to differentiate complicated functions by taking the natural logarithm (ln) of both sides first. It is especially helpful when the function involves:

• Products of many terms
• Quotients
• Variables raised to variable powers (like xx, (3x+1)sin⁡x)
• Functions that are hard to differentiate directly

**Method of solving Logarithmic Differentiation

• First, take the natural log on both sides of the equation given.
• Apply different properties of log to break the function and make it easier to solve.
• Differentiate the function by applying rules, like the chain rule.
• Multiply the RHS by the Function itself since it was in the denominator of the LHS.

**Derivative of logₐx (for any positive base a ≠ 1)

It is known that the differentiation of logx is 1/x, but this is the differentiation of natural log (that is, base e). Is it possible to have different bases, and can their differentiation be possible as well? YES. With the help of 2 simple properties of log, it can be derived.

Let's find out the derivative of Logax (where a is any positive integer a ≠ 1), d/dx(lnx)= 1/x

When the bases are changed, they can be written as, log_pq=\frac{logq}{logp}

Therefore, to writing logax in the form given above and then differentiating it will give, log_ax=\frac{logx}{loga}

Differentiating both sides,

d/dx[log_ax]=d/dx[\frac{logx}{loga}]\\=\frac{1}{loga}d/dx[logx]\\=\frac{1}{xloga}

**Example 1: Find the differentiation for log9x

**Solution:

d/dx[log9x]= d/dx[logx/log9]

d/dx[log_9x]=d/dx[\frac{logx}{log9}]\\=\frac{1}{log9}d/dx[logx]\\=\frac{1}{xlog9}

**Example 2: Differentiate -5log6x

**Solution:

d/dx[-5log_6x]=-5[d/dx[\frac{logx}{log6}]]\\=-5[\frac{1}{log6}d/dx[logx]]\\=\frac{-5}{xlog9}

**Example 3: Differentiate log4(x2+x)

**Solution:

As it is clear, that the function given is a Composite function. Therefore, chain rule is essential to be applied here.

y= log4(x2+x)

Assume x2+x be p(x)

p'(x)= 2x+1

For the log function, lets call it q(x)

q(x)= log4(x)

q'(x)=1/x.log4

y is the complete function which can now be written as,

y = q(p(x))

y' = q'(p(x))× p'(x)

dy/dx= \frac{1}{log4p(x)}p'(x)\\=\frac{1}{(x^2+x)log4}(2x+1)

Sample Problems on Logarithmic Differentiation

**Question 1: Differentiate, f(x)= log_2{(1-3x^3)}

**Solution:

Differentiating, f(x)= log_2{(1-3x^3)}

Assume, 1-3x3= v(x)

Where, v(x) is also a function of x, hence it is needed to be differentiated as well.

d/dx[f(x)]=d/dx[ log_2{(1-3x^3)}]\\=d/dx[ log_2{(v(x))}] \\=\frac{v'(x)}{log2{(v(x))}}\\= \frac{-9x^2}{log2(1-3x^3)}

**Question 2: Differentiate, h(x) = 5ln(x)

**Solution:

d/dx[ln(x)]= 1/(x)

Hence, d/dx[h(x)] = h'(x)= d/dx[5/x]

h'(x)= 5/x

**Question 3: Differentiate, y = ln(4+ 7x5)

**Solution:

y'= d[y]/dx =d/dx[ln(4+7x5)]

dy/dx= \frac{d/dx[4+7x^5]}{4+7x^5}

=\frac{35x^4}{4+7x^5}

**Question 4: Differentiate, y = cosx × cos3x × cos5x

**Solution:

Add log on both sides,

Logy = log{cosx × cos3x × cos5x}

Logy = log(cosx) × log(cos3x) × log(cos5x)

Differentiating on both sides,

d/dx[logy] = d/dx[log(cosx) × log(cos3x) × log(cos5x)]

1/y × dy/dx = [(1/cosx) × d(cosx)/dx] + [1/cos3x × d(cos3x)/dx] + [1/cos5x × d(cos5x)/dx]

1/y × dy/dx = -sinx/cosx -3sin3x/cos3x -5 sin5x/cos5x

dy/dx = y × {-tanx-3tan3x-5tan5x}

dy/dx = {cosx× cos3x × cos5x} × {-tanx -3tan3x -5tan5x}

**Question 5: What is the meaning of Log of a number?

**Answer:

A Log or Logarithms is the power to which a number must be raised in order to get another number.

**For example, the logarithm of base 10 for 1000 is 3, the base 10 logarithms of 10000 is 4, and so on. Log is used to find the skewness in large values and to show percent change of multiple factors.

**Question 7: Explain in steps to solve the Logarithmic Differentiation.

**Answer:

Steps to solve logarithmic differentiation are very easy and short,

• Take Log on both sides of the equation
• Use the properties of Log and simplify RHS
• Differentiate both sides, apply chain rule on RHS
• Put the value of the function both sides

**Question 8 : Differentiate, y=\frac{x+5}{x^3+3}

**Answer:

Apply log on both sides,

Logy= Log[\frac{x+5}{x^3+3}]

1/y. dy/dx= d/dx {log(x+5) -log(x3+ 3)}

1/y. dy/dx = 1/(x+5) - 3x/(x3+3)

dy/dx= y [1/(x+5) - 3x/(x3+3)]

dy/dx= \frac{x+5}{x^3+3}[\frac{1}{x+5}-\frac{3x}{x^3+3}]

**Question 9: Differentiate, y= \frac{(x+3)(x^5-2)}{(x+10)(x^2+1)}

**Answer:

y= \frac{(x+3)(x^5-2)}{(x+10)(x^2+1)}

Taking Log on both sides,

logy=log [\frac{(x+3)(x^5-2)}{(x+10)(x^2+1)}]

=log(x+3)+log(x^5-2)-log(x+10)-log(x^2+1)

\frac{dy}{dx}\frac{1}{y}=\frac{1}{x+3}+\frac{5x^4}{x^5-2}-\frac{1}{x+10}-\frac{2x}{x^2+1}

dy/dx=y[\frac{1}{x+3}+\frac{5x^4}{x^5-2}-\frac{1}{x+10}-\frac{2x}{x^2+1}]

dy/dx= [\frac{(x+3)(x^5-2)}{(x+10)(x^2+1)}][\frac{1}{x+3}+\frac{5x^4}{x^5-2}-\frac{1}{x+10}-\frac{2x}{x^2+1}]