Logarithmic differentiation
Use logarithms to turn products into sums and bring changing exponents down where you can differentiate them.
The idea
Products, quotients, and powers can become easier to differentiate after taking a logarithm. This is especially useful when both the base and the exponent depend on the input , as in \(x^x\).
The method combines logarithm laws with implicit differentiation.
Prerequisite
The logarithm rule
For differentiable \(u(x)>0\), the previous exponential and logarithmic lesson established
\[\frac{d}{dx}\ln(u(x))=\frac{u^{\prime}(x)}{u(x)}.\]
Use the product, quotient, and power laws of logarithms to rewrite positive factors before differentiating. A sum inside a logarithm stays intact.
Method
Logarithmic differentiation
On an interval where \(y=f(x)>0\):
Take \(\ln y\) on both sides.
Use logarithm laws to simplify the expression.
Differentiate with respect to \(x\), treating \(y\) as a function of \(x\).
Multiply by \(y\), then substitute the original expression for \(y\).
The left side becomes \(y^{\prime}/y\), not \(y^{\prime}\). Multiply by the original function to recover its derivative.
Worked examples
Example 1
A changing base and exponent
Let \(y=x^x\) for \(x>0\). The usual power rule assumes a constant exponent, so it does not apply directly. Instead,
\[\ln y=x\ln x.\]
Differentiate the product on the right:
\[\frac{y^{\prime}}y=\ln x+x\frac1x=\ln x+1.\]
Now multiply by \(y=x^x\):
\[y^{\prime}=x^x(\ln x+1).\]
Example 2
A product, a root, and a quotient
For \(x>1\), let
\[y=\frac{x^2\sqrt{x-1}}{(x+2)^3}.\]
All factors are positive on this interval, so
\[\ln y=2\ln x+\frac12\ln(x-1)-3\ln(x+2).\]
Differentiate each term, including its inner derivative:
\[\frac{y^{\prime}}y=\frac2x+\frac1{2(x-1)}-\frac3{x+2}.\]
Thus
\[y^{\prime}=\frac{x^2\sqrt{x-1}}{(x+2)^3}\left(\frac2x+\frac1{2(x-1)}-\frac3{x+2}\right).\]
The factored answer keeps the original function visible and avoids expanding a large numerator.
Example 3
A variable power with an inner function
For \(x>-1\), let \(y=(1+x)^{\sin x}\). The base is positive, so
\[\ln y=\sin(x)\cdot\ln(1+x).\]
Apply the product rule and chain rule:
\[\frac{y^{\prime}}y=\cos(x)\cdot\ln(1+x)+\frac{\sin x}{1+x}.\]
Multiply by the original function:
\[y^{\prime}=(1+x)^{\sin x}\left(\cos(x)\cdot\ln(1+x)+\frac{\sin x}{1+x}\right).\]
Both changing parts contribute: one term comes from the exponent, and one from the base.
Warning
Logarithms do not split sums: \(\ln(u+v)\ne\ln u+\ln v\) in general. Do not take the logarithm of zero, and do not stop at \(y^{\prime}/y\) when the question asks for \(y^{\prime}\). If the original function has a zero, check its derivative there by another method.
Quiz
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For \(y=(x^2+1)^3\), which logarithmic rewrite is valid?
\(\ln y=6\ln x\) \(\ln y=3\ln(x^2+1)\) \(\ln y=3(\ln(x^2)+\ln1)\) Explanation The power becomes a multiplier, but the sum inside remains intact. There is no rule that splits the logarithm of a sum into a sum of logarithms.For \(y=2^x\), taking logarithms gives \(\ln y=x\ln2\). What follows after differentiating?
\(y^{\prime}/y=\ln2\) \(y^{\prime}/y=x/2\) \(y^{\prime}/y=1\) \(y^{\prime}=x\ln2\) Explanation The number \(\ln2\) is constant, so the derivative of \(x\ln2\) is \(\ln2\). Multiplying by \(y\) then gives \(y^{\prime}=2^x\ln2\).A positive differentiable function has \(f(2)=10\) and \(f^{\prime}(2)=3\). What is the slope of \(\ln(f(x))\) at \(x=2\)?
\(3\) \(30\) \(0.3\) \(\ln3\) Explanation The chain rule gives \(d[\ln(f(x))]/dx=f^{\prime}(x)/f(x)\). At this input, the ratio is \(3/10=0.3\), not the logarithm of the original slope.For \(y=x-1\), logarithmic differentiation is unavailable at \(x=1\). Does that mean the original function has no derivative there?
Yes: zero values prevent differentiability No: its derivative is \(1\); only the logarithmic method fails there No: its derivative is \(0\) there Explanation The original function is a line with slope one. At its zero, \(\ln y\) and \(y^{\prime}/y\) are undefined, so use another method.For \(y=\sqrt{2-x}/(x-1)\), which interval lets us take the logarithm of each positive factor and of \(y\)?
\(x<1\) \(1<x<2\) \(x>2\) \(1<x\le2\) Explanation We need \(2-x>0\) and \(x-1>0\), giving \(1<x<2\). At \(x=2\), the original function is zero, so its logarithm is unavailable.A positive function satisfies \(\frac{d}{dx}\ln(y)=0\) at one input. What follows at that input?
\(y=1\) \(y^{\prime}=0\) \(y=0\) \(y\) is constant everywhere Explanation Since \(\frac{d}{dx}\ln(y)=y^{\prime}/y\) and \(y>0\), a zero logarithmic derivative means a zero ordinary derivative there. It determines neither the height nor the behavior at other inputs.
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