Logarithmic Differentiation Calculator

1. Logarithmic Differentiation Calculator Emathhelp
2. Use The Properties Of Logarithms Calculator

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. Anti-logarithm calculator. In order to calculate log-1 (y) on the calculator, enter the base b (10 is the default value, enter e for e constant), enter the logarithm value y and press the = or calculate button. Logarithmic differentiation will provide a way to differentiate a function of this type. It requires deft algebra skills and careful use of the following unpopular, but well-known, properties of logarithms. Though the following properties and methods are true for a logarithm of any base. Solve Exponential and logarithmic functions problems with our Exponential and logarithmic functions calculator and problem solver. Get step-by-step solutions to your Exponential and logarithmic functions problems, with easy to understand explanations of each step. General derivatives. Free Logarithmic Form Calculator - present exponents in their logarithmic forms step-by-step This website uses cookies to ensure you get the best experience. By using this website, you agree to our Cookie Policy.

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. True fear: forsaken souls part 1 for mac os. This approach allows calculating derivatives of power, rational and some irrational functions in an efficient manner.

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)}},]

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.

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).}]

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),}]

Logarithmic Differentiation Calculator Emathhelp

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.}]

File size 160.22 KB Mime type Stdin has more than one entry-rest ignored compressed-encoding=application/zip; charset=binary Other. This war of mine: the little ones crack.

Further we differentiate the left and right sides:

[
{{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 The Properties Of Logarithms Calculator