Logarithmic Differentiation Calculator

1. Derivative Calculator
2. Implicit Differentiation Calculator
3. Use The Properties Of Logarithms Calculator
4. Logarithmic Differentiation Calculator Mathway
5. Use Logarithmic Differentiation

The method of differentiating functions by first taking logarithms and then differentiating is called logarithmic differentiation. We use logarithmic differentiation in situations where it is easier to differentiate the logarithm of a function than to differentiate the function itself. This approach allows calculating derivatives of power, rational and some irrational functions in an efficient manner.

Calculate chain rule of derivatives with napierian logarithm If u is a differentiable function, the chain rule of derivatives with the napierian logarithm function and the function u is calculated using the following formula: (ln(u(x))'=`(u'(x))/(u(x))`, the derivative calculator can perform this type of calculation as this example shows. This free log calculator solves for the unknown portions of a logarithmic expression using base e, 2, 10, or any other desired base. Learn more about log rules, or explore hundreds of other calculators addressing topics such as math, finance, health, and fitness, among others.

Consider this method in more detail. Let (y = fleft( x right)). Take natural logarithms of both sides:

[ln y = ln fleft( x right).]

Next, we differentiate this expression using the chain rule and keeping in mind that (y) is a function of (x.)

[
{{left( {ln y} right)^prime } = {left( {ln fleft( x right)} right)^prime },;;}Rightarrow
{frac{1}{y}y'left( x right) = {left( {ln fleft( x right)} right)^prime }.}
]

It's seen that the derivative is

[
{y' = y{left( {ln fleft( x right)} right)^prime } }
= {fleft( x right){left( {ln fleft( x right)} right)^prime }.}
]

The derivative of the logarithmic function is called the logarithmic derivative of the initial function (y = fleft( x right).)

This differentiation method allows to effectively compute derivatives of power-exponential functions, that is functions of the form

[y = u{left( x right)^{vleft( x right)}},]

Derivative Calculator

where (uleft( x right)) and (vleft( x right)) are differentiable functions of (x.)

In the examples below, find the derivative of the function (yleft( x right)) using logarithmic differentiation.

Solved Problems

Click or tap a problem to see the solution.

Solution.

Implicit Differentiation Calculator

First we take logarithms of the left and right side of the equation:

[
{ln y = ln {x^x},;;}Rightarrow
{ln y = xln x.}
]

Now we differentiate both sides meaning that (y) is a function of (x:)

[
{{left( {ln y} right)^prime } = {left( {xln x} right)^prime },;;}Rightarrow
{frac{1}{y} cdot y' = x'ln x + x{left( {ln x} right)^prime },;;}Rightarrow
{frac{{y'}}{y} = 1 cdot ln x + x cdot frac{1}{x},;;}Rightarrow
{frac{{y'}}{y} = ln x + 1,;;}Rightarrow
{y' = yleft( {ln x + 1} right),;;}Rightarrow
{y' = {x^x}left( {ln x + 1} right),;;}kern0pt{text{where};;x gt 0.}
]

Solution.

First we take logarithms of both sides:

[{ln y = ln {x^{frac{1}{x}}},};; Rightarrow {ln y = frac{1}{x}ln x.}]

Differentiate the last equation with respect to (x:)

[{left( {ln y} right)^prime = left( {frac{1}{x}ln x} right)^prime,};; Rightarrow {frac{1}{y} cdot y^prime = left( {frac{1}{x}} right)^primeln x + frac{1}{x}left( {ln x} right)^prime,};; Rightarrow {frac{{y^prime}}{y} = – frac{1}{{{x^2}}} cdot ln x + frac{1}{x} cdot frac{1}{x},};; Rightarrow {frac{{y^prime}}{y} = frac{1}{{{x^2}}}left( {1 – ln x} right),};; Rightarrow {y^prime = frac{y}{{{x^2}}}left( {1 – ln x} right).}]

Substitute the original function instead of (y) in the right-hand side:

[{y^prime = frac{{{x^{frac{1}{x}}}}}{{{x^2}}}left( {1 – ln x} right) }={ {x^{frac{1}{x} – 2}}left( {1 – ln x} right) }={ {x^{frac{{1 – 2x}}{x}}}left( {1 – ln x} right).}]

Use The Properties Of Logarithms Calculator

Solution.

Apply logarithmic differentiation:

[
{ln y = ln left( {{x^{ln x}}} right),;;}Rightarrow
{ln y = ln xln x = {ln ^2}x,;;}Rightarrow
{{left( {ln y} right)^prime } = {left( {{{ln }^2}x} right)^prime },;;}Rightarrow
{frac{{y'}}{y} = 2ln x{left( {ln x} right)^prime },;;}Rightarrow
{frac{{y'}}{y} = frac{{2ln x}}{x},;;}Rightarrow
{y' = frac{{2yln x}}{x},;;}Rightarrow
{y' = frac{{2{x^{ln x}}ln x}}{x} }={ 2{x^{ln x – 1}}ln x.}
]

Solution.

Take the logarithm of the given function:

[
{ln y = ln left( {{x^{cos x}}} right),;;}Rightarrow
{ln y = cos xln x.}
]

Differentiating the last equation with respect to (x,) we obtain:

[
{{left( {ln y} right)^prime } = {left( {cos xln x} right)^prime },;;}Rightarrow
{frac{1}{y} cdot y' }={ {left( {cos x} right)^prime }ln x + cos x{left( {ln x} right)^prime },;;}Rightarrow
{{frac{{y'}}{y} }={ left( { – sin x} right) cdot ln x + cos x cdot frac{1}{x},;;}}Rightarrow
{{frac{{y'}}{y} }={ – sin xln x + frac{{cos x}}{x},;;}}Rightarrow
{{y' }={ yleft( {frac{{cos x}}{x} – sin xln x} right).}}
]

Substitute the original function instead of (y) in the right-hand side:

[{y' = {x^{cos x}}cdot}kern0pt{left( {frac{{cos x}}{x} – sin xln x} right),}]

where (x gt 0.)

Solution.

Taking logarithms of both sides, we get

[{ln y = ln {x^{arctan x}},};; Rightarrow {ln y = arctan xln x.}]

Differentiate this equation with respect to (x:)

[{left( {ln y} right)^prime = left( {arctan xln x} right)^prime,};; Rightarrow {frac{1}{y} cdot y^prime = left( {arctan x} right)^primeln x }+{ arctan xleft( {ln x} right)^prime,};; Rightarrow {frac{{y^prime}}{y} = frac{1}{{1 + {x^2}}} cdot ln x }+{ arctan x cdot frac{1}{x},};; Rightarrow {frac{{y^prime}}{y} = frac{{ln x}}{{1 + {x^2}}} }+{ frac{{arctan x}}{x},};; Rightarrow {y^prime = yleft( {frac{{ln x}}{{1 + {x^2}}} + frac{{arctan x}}{x}} right),}]

where (y = {x^{arctan x}}.)

Solution.

Taking logarithms of both sides, we can write the following equation:

[{ln y = ln {x^{2x}},;;} Rightarrow {ln y = 2xln x.}]

Further we differentiate the left and right sides:

Logarithmic Differentiation Calculator Mathway

[
{{left( {ln y} right)^prime } = {left( {2xln x} right)^prime },;;}Rightarrow
{frac{1}{y} cdot y' }={ {left( {2x} right)^prime } cdot ln x + 2x cdot {left( {ln x} right)^prime },;;}Rightarrow
{frac{{y'}}{y} = 2 cdot ln x + 2x cdot frac{1}{x},;;}Rightarrow
{frac{{y'}}{y} = 2ln x + 2,;;}Rightarrow
{y' = 2yleft( {ln x + 1} right);;}kern0pt{text{or};;y' = 2{x^{2x}}left( {ln x + 1} right).}
]

Use Logarithmic Differentiation