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Logarithmic Differentiation

April 9th, 2009 |

Author: Merrin

Taking their derivatives of certain functions can be greatly simplified by working with natural logarithms. The main properties of natural logarithms that we will frequently use with differentiation are

$\ln A^{B} =B\ln A$

$\ln \left(\cfrac{AB}{CD} \right)=\ln A+\ln B-\ln C-\ln D$

Also, we know the derivative of the natural logarithm is given by

$\cfrac{d}{dx} \ln x=\cfrac{1}{x}$

By the chain rule, we have the derivative of the natural logarithm of a more general function.

$\cfrac{d}{dx} \ln f(x)=\cfrac{f'(x)}{f(x)}$

This equation can be rearranged to form

$f'(x)=f(x)\cfrac{d}{dx} \ln f(x)$

This is the main formula of logarithmic differentiation. Now depending on what the function f(x) is we will use a property of the logarithm law on the right hand side of this equation to carry out the simplification. In general, the types of functions that are simplified by taking logarithms are products of functions or iterated exponential functions where the above properties of the logarithm are useful.

Example 1. Prove the product rule for three functions

$\cfrac{d}{dx} (uvw)=?$

Solution 1. Since the logarithm of a product of functions can be split into a sum, we will use logarithmic differentiation.

$\ln (uvw)=\ln f=\ln u+\ln v+\ln w$

Now we differentiate both sides of the equation and we have

${\cfrac{f'(x)}{f(x)} } = {\cfrac{u'}{u} +\cfrac{v'}{v} +\cfrac{w'}{w} } \\ {f'(x)} = {f(x)\left(\cfrac{u'}{u} +\cfrac{v'}{v} +\cfrac{w'}{w} \right)=uvw\left(\cfrac{u'}{u} +\cfrac{v'}{v} +\cfrac{w'}{w} \right)} \\ {f'(x)} = {u'vw+uv'w+uvw'}$

This problem is also solvable by applying the product rule directly twice.

$\cfrac{d}{dx} [(u)(vw)]=(u)'(vw)+(u)(vw)'=u'vw+uv'w+uvw'$

Example 2. Generalize the triple product rule and use it to derive the power rule again a different way.

Solution 2. From the previous expressions we can directly see how to make a product rule for n functions which is such that every function in the product gets a turn to have its derivative.

$\cfrac{d}{dx} \prod _{i=1}^{n}g_{i} (x)=\sum _{j=1}^{n} g'_{j} (x) \prod _{\begin{array}{l} {k=1} \\ {k\ne j} \end{array}}^{n}g_{k} (x)$

We can apply this rule. Let us take the derivative of n factors of x so we can derive the power rule from the n product rule.

$\cfrac{d}{dx} x^{n} =x'xx...+xx'x...+...=nx^{n-1} x'=nx^{n-1} \quad \quad n\in {\mathbb N}$

Example 3. Logarithmic differentiation is a form of implicit differentiation. Take the derivative of the following function by logarithmic differentiation.

$y=a^{x}$

Solution 3. First we take the log.

$\ln y=\ln a^{x} =x\ln a$

Now we take the derivative of both sides of the equation.

$\cfrac{y'}{y} =\ln a$

$y'=y\ln a$

In this case, we already know y explicitly as a function of x so we can substitute that back in.

$\cfrac{dy}{dx} =a^{x} \ln a$

Logarithmic differentiation is useful is to chop down small towers of repeated exponentials.

Example 4. Find the derivative of

$f(x)=x^{x}$

Solution 4. We first take the logarithm of both sides, apply a property of the logarithm, and then differentiate.

${\cfrac{d}{dx} \ln f=\cfrac{d}{dx} (\ln x^{x} )=\cfrac{d}{dx} (x\ln x)} \\ {\cfrac{f'}{f} =\ln x+1} \\ {f'=f(\ln x+1)} \\ {\cfrac{d}{dx} x^{x} =x^{x} (\ln x+1)}$

Example 5. Here is a similar but slightly harder example. Find the derivative of

$f(x)=(\ln x)^{(\ln x)}$

Solution 5. We see a repeated exponential so a good place to start is with logarithmic differentiation.

$\ln f(x)=\ln (\ln x)^{(\ln x)} =(\ln x)\ln (\ln x)$

Now we can use the product rule in differentiating the right hand side.

${\cfrac{f'}{f} =\cfrac{d}{dx} [(\ln x)\ln (\ln x)]=\cfrac{1}{x} \ln (\ln x)+(\ln x)\cfrac{1}{\ln x} \cfrac{1}{x} } \\ {f'=\cfrac{f}{x} (\ln (\ln x)+1)=\cfrac{(\ln x)^{(\ln x)} }{x} (\ln (\ln x)+1)}$

Example 6. There is a general formula to these problems with repeated exponentials that we can derive.

$\cfrac{d}{dx} g(x)^{h(x)} =?$

Solution 6. This generalization can be solved with logarithmic differentiation

${\ln f=\ln g^{h} =h\ln g} \\ {\cfrac{f'}{f} =h'\ln g+\cfrac{hg'}{g} } \\ {f'=g^{h} \left(h'\ln g+\cfrac{hg'}{g} \right)}$

There is another form of this expression which I particularly like which leads to a general pattern for derivatives of higher towers.

$f'=\cfrac{d}{dx} g^{h} =g^{h} h\left((\ln h)'\ln g+(\ln g)'\right)$

Example 7. Prove the power rule for real n.

Solution 7. We set up our function, take the natural logarithm, and then differentiate both sides of the equation.

$f(x)=x^{n}$

$\log f(x)=\log (x^{n} )=n\log x$

$\cfrac{f'(x)}{f(x)}=n\cfrac{1}{x}$

$\cfrac{d}{dx} x^{n} =n\cfrac{f(x)}{x} =n\cfrac{x^{n} }{x} =nx^{n-1} \quad n\in {\mathbb R}$

Example 8. Prove using logarithmic differentiation that

$\cfrac{d}{dx} e^{e^{x} } =e^{e^{x} } c^{x}$

Solution 8. We apply the same recipe again.

${f(x)=e^{e^{x} } } \\\\ {\ln f=e^{x} \ln e=e^{x} } \\\\ {\cfrac{f'}{f} =\cfrac{d}{dx} e^{x} =e^{x} } \\\\ {f'=fe^{x} } \\\\ {\cfrac{d}{dx} e^{e^{x} } =e^{e^{x} } e^{x} }$

Example 9. Find the derivative of

$f(x)=\cfrac{(1-x^{2} )^{2} \sqrt{x} (\text{arcsin}\, x)^{4} }{a^{x} (\cos x)^{3} }$

Solution 9. If there is a monstrous product of functions then logarithmic differentiation is probably essential. Notice that factors in the denominator are just like factors in the numerator raised to a negative power. In any case, split up the function into pieces by taking the logarithm then take the derivative.

It is actually quite easy to find the derivatives of certain monsters by knowing all your fundamental derivatives and how to take a logarithm.

$\ln f(x)=2\ln (1-x^{2} )+0.5\ln x+4\ln (\text{arcsin}\, x)-x\ln a-3\ln \cos x$

$\cfrac{f'}{f} =2\cfrac{1}{1-x^{2} } (-2x)+\cfrac{1}{2x} +4\cfrac{1}{\text{arcsin}\, x} \cfrac{1}{\sqrt{1-x^{2} } } -\ln a-3\cfrac{1}{\cos x} (-\sin x)$

$\cfrac{f'}{f} = \cfrac{-4x}{1-x^{2} } +\cfrac{1}{2x} +\cfrac{4}{\text{arcsin}\, x} \cfrac{1}{\sqrt{1-x^{2} } } -\ln a+3\tan x$

$\cfrac{d}{dx}f(x)=f(x)\left(\cfrac{-4x}{1-x^{2} } +\cfrac{1}{2x} +\cfrac{4}{\text{arcsin}\, x} \cfrac{1}{\sqrt{1-x^{2} } } -\ln a+3\tan x\right)$

Example 10. Find the derivative of the following expression.

$f(x)=g_{1} (x)^{p_{1} } g_{2} (x)^{p_{2} } ...$

Solution 10. This is actually quite a complicated and general expression. There are products of arbitrary functions that are exponentiated and may occur in the numerator with positive powers and in the denominator with negative powers. The nature of the derivative of products is revealed when we take the derivative by logarithmic differentiation. This procedure helps differentiate monsters.

${\ln f=p_{1} \ln g_{1} (x)+p_{2} \ln g_{2} (x)+...} \\ {\cfrac{f'}{f} =p_{1} \cfrac{g_{1} ^{{'} } (x)}{g_{1} (x)} +p_{2} \cfrac{g_{2} ^{{'} } (x)}{g_{2} (x)} +...} \\ {\cfrac{d}{dx} f(x)=p_{1} g'_{1} (x)g_{2} (x)...+p_{2} g_{1} (x)g'_{2} (x)...} \\ {\cfrac{d}{dx} \prod _{i=1}^{n}g_{i} (x)=\sum _{i=1}^{n}p_{i} g'_{i} (x) \prod _{j=1}^{n}g_{j}(x)}$