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.

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.