Logarithmic Derivatives: From Slope to a Powerful Method

The slope of ln x is 1/x

At a positive input x, the derivative measures the limiting slope of nearby secants on $y=\ln x$. Begin with the definition:

For sufficiently small h, both x and x+h are positive. Combine the logarithms, then divide the small change by x:

As h approaches zero, h/x also approaches zero. The bracketed logarithm ratio tends to 1, giving $$\boxed{\frac{d}{dx}\ln x=\frac1x\qquad(x>0).}$$

The formula agrees with the graph. Its slope is positive everywhere in its domain, so the curve increases. But the slope becomes smaller as x grows: at x=1 it is 1, at x=2 it is 1/2, and at x=5 it is 1/5.

A tangent slope is not the height of the curve. At x=1, the height is $\ln1=0$, while the slope is 1. Keep these quantities separate when reading the exploration. To find a tangent line, use both the point and the slope in $y-y_0=m(x-x_0)$.

There is no tangent to the real ln curve at x=0 because that point is outside the domain. The derivative formula does not extend the original function to nonpositive inputs.

Other bases keep a constant factor

Changing the logarithm base changes a constant multiplier. For a fixed $b>0$, $b\ne1$, $$\log_bx=\frac{\ln x}{\ln b}.$$ Since $\ln b$ does not depend on x, differentiation gives $$\boxed{\frac{d}{dx}\log_bx=\frac1{x\ln b}\qquad(x>0).}$$

If b>1, the factor ln b is positive and the derivative is positive. If 0<b<1, ln b is negative and the derivative is negative. This is the calculus version of the two logarithm graph families. For the common logarithm, $d(\log_{10}x)/dx=1/(x\ln10)$, not simply 1/x.

Exponentials have a matching pair of rules. The standard limit for $(a^h-1)/h$ gives $$\frac{d}{dx}a^x=a^x\ln a\qquad(a>0).$$ In particular, $d(e^x)/dx=e^x$, because ln e=1. If a=1, the function is constant and the derivative is 0; the same expression gives 1·ln 1=0.

The multiplier ln a is essential. For example, the slope of $2^x$ is $2^x\ln2$, whereas the slope of $(1/2)^x$ is negative. Differentiate the function that is actually written rather than treating every exponential as e^x.

In these rules the base is a fixed constant. If the base itself depends on x, as in x^x, a different argument is needed. We will build that method after the chain rule.

Differentiate the inside as well

For $y=\ln(3x-2)$, the logarithm is applied to an expression that changes three times as fast as x. The chain rule records both layers: differentiate the outer logarithm, then multiply by the derivative of its argument.

This rule requires f to be differentiable and $f(x)>0$ on the interval being considered. The positivity condition belongs to the original logarithm; it cannot be discarded after simplifying the derivative.

For a fixed valid base b, the extra constant remains:

Similarly, $d(e^{f(x)})/dx=e^{f(x)}f'(x)$, and $d(a^{f(x)})/dx=a^{f(x)}\ln a\,f'(x)$ for a fixed a>0. Always identify the inside function before applying a memorised rule.

Compare $\ln(x^2+1)$ and $\ln(x^2-1)$. Their derivatives look similar, but their domains differ. The first exists for every real x and has derivative $2x/(x^2+1)$. The second requires $x<-1$ or $x>1$, and has derivative $2x/(x^2-1)$ only on those intervals.

A useful check is to name the inside, its derivative and its allowed inputs in three short steps. This prevents both a missing multiplier and a derivative used where the original function never existed.

Use logs to untangle a positive product

A long product with powers may be easier to differentiate after taking a logarithm. Logarithms turn products into sums, quotients into differences and powers into coefficients. Differentiate the simpler relation, then recover the derivative of the original function.

Consider $$y=\frac{x^2(x+1)^3}{x+2},\qquad x>0.$$ On this stated interval, every factor and y itself is positive. Taking natural logarithms is therefore legal:

Now differentiate both sides. The left side also needs the chain rule because y depends on x:

Finally multiply by y and substitute its original expression:

Stopping at y′/y would give the relative rate of change, not the derivative y′. The last multiplication is part of the method.

Why did we state x>0? It makes the positive-log expansion valid in every line. The original rational expression may exist elsewhere, but this derivation is deliberately restricted to the stated interval. If a problem asks for other sign intervals, handle them carefully with absolute-value logarithms or direct differentiation; do not take ln of a negative factor.

Choose this method because it simplifies the structure, not because every product must be treated this way. A short product may be quicker with the ordinary product rule.

When both the base and exponent change

In $y=x^x$, neither the ordinary fixed-power rule nor the fixed-base exponential rule applies on its own: x appears in both positions. Work on x>0 so that the real variable power can be written as $e^{x\ln x}$.

Taking logarithms exposes a familiar product:

Differentiate with respect to x. The left side gives y′/y; the right side uses the product rule:

Multiplying by y gives the result $$\boxed{\frac{d}{dx}x^x=x^x(\ln x+1),\qquad x>0.}$$

The derivative’s sign is especially easy to read. The factor x^x is positive, so the sign comes from ln x + 1. It is negative for 0<x<1/e, zero at x=1/e and positive for x>1/e. Thus the function decreases and then increases, with a minimum at 1/e. A changing exponent does not force a function to increase everywhere.

The same method handles $y=f(x)^{g(x)}$ on an interval where f>0 and both functions are differentiable. Write $\ln y=g\ln f$, then use the product and chain rules:

Use the expanded reasoning until the two contributions are clear: one comes from the changing exponent, the other from the changing base. Keep the positive-base interval visible throughout.