Establish and use the derivative of the natural logarithm function, including the chain rule for the logarithm of a function of x

What this dot point is asking

SCSA Unit 3 treats the natural logarithm $\ln x$ as the inverse of $e^{x}$, and this dot point asks you to differentiate it and any composite logarithm. The result is examined in both examination sections and underpins later integration of $\dfrac{1}{x}$.

The derivative of $\ln x$

Because $\ln x$ is the inverse of $e^{x}$, write $y=\ln x$ so that $x=e^{y}$. Differentiating $x=e^{y}$ with respect to $x$ gives

So the gradient of $\ln x$ at any point is the reciprocal of the $x$-coordinate.

The chain rule with logarithms

For a logarithm of a function, the chain rule divides the derivative of the inside by the inside itself. If $y=\ln(f(x))$ with outer $\ln u$ and inner $u=f(x)$, then $\dfrac{dy}{dx}=\dfrac{1}{u}\cdot u'=\dfrac{f'(x)}{f(x)}$.

Simplify first with log laws

Logarithm laws let you turn products, quotients and powers into sums, differences and multiples before differentiating, which avoids the quotient rule entirely.

Logarithms inside products and quotients

A logarithm often appears as a factor in a product, where the product rule combines with $\dfrac{d}{dx}\ln x=\dfrac{1}{x}$. The result $\dfrac{d}{dx}(x\ln x)=\ln x+1$ is worth recognising: it shows that the antiderivative of $\ln x$ is $x\ln x-x$, a result used later in integration. When a logarithm sits in the numerator or denominator of a quotient, expanding with log laws first usually avoids the quotient rule entirely, as in the worked expansion above.

Tangents and rates with logarithms

Because $\dfrac{d}{dx}\ln x=\dfrac{1}{x}$ is large near $x=0^{+}$ and small for large $x$, the logarithm curve is steep at first then flattens. This is why $\ln x$ models diminishing-returns growth. To find a tangent, evaluate $\dfrac{1}{x}$ at the point for the gradient, then use $y-y_{1}=m(x-x_{1})$. For example, the tangent to $y=\ln x$ at $x=1$ has gradient $\dfrac{1}{1}=1$ and passes through $(1,0)$, giving $y=x-1$.

A logarithm inside a chain rule

When a logarithm is composed with another function, layer the rules. For $y=\bigl(\ln x\bigr)^{3}$, treat it as an outer power with inner $\ln x$:

The factor $\dfrac{1}{x}$ comes from differentiating the inner $\ln x$. Contrast this with $\ln(x^{3})$, which log laws simplify to $3\ln x$ with derivative $\dfrac{3}{x}$. Reading whether the power is inside or outside the logarithm changes the method entirely.

Base-$a$ logarithms

For a logarithm to base $a$, use the change of base $\log_{a}x=\dfrac{\ln x}{\ln a}$, so $\dfrac{d}{dx}\log_{a}x=\dfrac{1}{x\ln a}$. The constant $\ln a$ is just a multiplier. SCSA Methods calculus almost always uses the natural logarithm.