Why rewrite radicals and reciprocals?
Some functions look complicated because the same factors appear in different forms: under radicals, in denominators, or both. Rewriting these pieces as powers gives a single format that logarithmic differentiation can handle systematically.
For example, a square root is a power with exponent \(1/2\), while a factor in a denominator can be written with a negative exponent. Once the function is expressed as a product of powers, taking its logarithm turns products into sums. Differentiating that sum is often more straightforward than applying the product and quotient rules repeatedly.
The method works on an interval where the function is positive and the factors inside the logarithms are positive. These conditions matter: the real logarithm \(\ln u\) is defined only when \(u>0\).
Rewrite the factors as powers
For a positive quantity \(u\), a radical can be expressed using a fractional exponent. A reciprocal can be expressed using a negative exponent. For instance, when \(u>0\), \(\sqrt{u}=u^{1/2}\), and \(1/u^2=u^{-2}\). More generally, a positive \(n\)th root can be written as a power with exponent \(1/n\).
The positive-factor condition lets us take the logarithm of every \(u_j(x)\). The function \(y\) is then positive as well, even when some exponents are negative. In applications, first identify an interval where the original expression and its power form agree and all the factors being logged are positive.
The logarithmic derivative for powers
For the expression in the definition, take the natural logarithm of both sides. The logarithm of a product is the sum of the logarithms, and the logarithm of a positive base raised to a power pulls down that exponent.
Differentiate both sides with respect to \(x\). The chain rule gives \(\frac{d}{dx}\ln y=y'/y\) and \(\frac{d}{dx}\ln u_j(x)=u_j'(x)/u_j(x)\). Thus:
Multiplying by \(y\) gives the derivative. This identity explains the method: each factor contributes its exponent times its own relative derivative, and the original function multiplies the total.
A reliable procedure
Keep the original function visible while differentiating its logarithm. The logarithmic derivative \(y'/y\) is not itself the derivative \(y'\); the final multiplication by \(y\) is essential.
Check where the original expression is defined and where the factors to be logged are positive.
Use fractional exponents for radicals and negative exponents for denominator factors.
Write \(\ln y\) as a sum of exponent-weighted logarithms.
Find \(y'/y\), then multiply by the original power-form expression \(y\).
Worked examples
Worked Example: A square root divided by a squared factor
Differentiate \(y=\frac{\sqrt{x+1}}{x^2}\) for \(x>0\). On this interval, \(x+1>0\) and \(x>0\), so both factors can be logged. Rewrite the function as \(y=(x+1)^{1/2}x^{-2}\).
Take logarithms and differentiate:
Put the terms over a common denominator:
Multiply by \(y=(x+1)^{1/2}x^{-2}\):
To check, differentiate the power form using the product rule:
With denominator \(2x^3\sqrt{x+1}\), the numerator is \(x-4(x+1)=-3x-4\), giving the same derivative. As a numerical check at \(x=1\), the formula gives \(y'=-7\sqrt{2}/4\); the product-rule expression gives \(\sqrt{2}/4-2\sqrt{2}=-7\sqrt{2}/4\) as well.
Worked Example: Several roots and a reciprocal power
For \(x>1\), differentiate \(y=\frac{\sqrt{x-1}\sqrt[4]{x+2}}{x^2\sqrt{x+3}}\). All four bases are positive on this interval. Rewrite each radical and denominator factor:
Taking logarithms gives:
Differentiate each term. For example, \(\frac{d}{dx}\ln(x-1)=1/(x-1)\); the negative exponent on \(x\) contributes \(-2/x\).
Therefore the derivative, retaining the original function as a factor, is:
Check the value at \(x=2\). The function is \(y(2)=\sqrt{2}/(4\sqrt{5})\), and the logarithmic derivative is \(\frac12+\frac1{16}-1-\frac1{10}=-\frac{43}{80}\). Hence \(y'(2)=-43\sqrt{2}/(320\sqrt{5})\).
For a direct check, differentiate the four-factor power form by the product rule and factor out \(y\). The four terms left inside the brackets are respectively \(\frac{1}{2(x-1)}\), \(\frac{1}{4(x+2)}\), \(-\frac{2}{x}\), and \(-\frac{1}{2(x+3)}\). Thus the product-rule derivative is exactly the expression above, including its value at \(x=2\).
Worked Example: A radical of a quotient with a reciprocal root
For \(x>1\), differentiate \(y=\sqrt{\frac{x^2+1}{(x-1)^3}}\frac{1}{\sqrt[3]{x+2}}\). The quantities \(x^2+1\), \(x-1\), and \(x+2\) are positive on this interval. Separate the factors and rewrite the expression:
After taking logarithms:
Use the chain rule on each logarithm. In particular, the derivative of \(\ln(x^2+1)\) is \(2x/(x^2+1)\), so its contribution after multiplication by \(1/2\) is \(x/(x^2+1)\).
Multiplying by \(y\) gives:
Check at \(x=2\). Here \(y(2)=\sqrt{5}/\sqrt[3]{4}\), and \(\frac{y'(2)}{y(2)}=\frac25-\frac32-\frac1{12}=-\frac{71}{60}\). Therefore \(y'(2)=-71\sqrt{5}/(60\sqrt[3]{4})\).
For a direct derivative check, apply the product rule to the three power factors. After factoring out \(y\), their respective contributions are \(\frac12\frac{2x}{x^2+1}=\frac{x}{x^2+1}\), \(-\frac32\frac{1}{x-1}\), and \(-\frac13\frac{1}{x+2}\). These sum to the displayed logarithmic derivative, so the direct derivative agrees, including at \(x=2\).
A common sign error
A reciprocal contributes a negative exponent, and that negative sign must remain when taking logarithms. For example, for \(y=\sqrt{x+1}/x^2\), a mistaken step would be:
This treats the denominator as though it were a numerator. The correct rewrite is \(y=(x+1)^{1/2}x^{-2}\), so the correct logarithm is:
Consequently, the contribution from \(x^{-2}\) to \(y'/y\) is \(-2/x\), not \(+2/x\). A useful check is to ask whether every factor originally in the denominator still carries a negative exponent after rewriting.
Why the method is useful
With several radical and reciprocal factors, a direct product or quotient rule can create many terms, each of which must still be simplified. Logarithmic differentiation organizes the calculation factor by factor: the exponent becomes a coefficient, and the derivative of each positive base contributes its derivative divided by the base. The original function is then multiplied back at the end.
This organization does not remove the need to check the domain, apply the chain rule, or handle signs accurately. It provides a compact route to the derivative when a positive function is naturally written as a product of powers. In the examples, differentiating the power form directly confirms that the logarithmic calculation preserves every factor and sign.
Check Your Understanding
For each question, first write the expression using powers. Keep track of the interval on which the bases being logged are positive.
- Rewrite \(\frac{\sqrt{x+4}}{x^3}\) as a product of powers.
- For \(y=(x+1)^{1/2}(x+3)^{-2}\), what is \(y'/y\)?
- Why does the factor \(1/(x-2)^3\) contribute \(-3/(x-2)\) to the logarithmic derivative on an interval where \(x-2>0\)?
- For \(y=\sqrt{x^2+1}/x\) on \(x>0\), find \(y'/y\).
- What final step is required after finding \(y'/y\) in order to obtain \(y'\)?
