Mathematics 9709/23 — October/November 2011

Walkthrough

We need the gradient of the curve at a point, so we differentiate $y$ with respect to $x$. Since $y = \ln(5x+1)$ is a natural logarithm of a linear expression, we use the chain rule. Write $u = 5x+1$; then $\frac{du}{dx} = 5$ and $\frac{d}{du}(\ln u) = \frac{1}{u}$. Multiplying gives $\frac{dy}{dx} = \frac{5}{5x+1}$. Then substitute $x = 4$: $\frac{dy}{dx} = \frac{5}{5(4)+1} = \frac{5}{21}$. The gradient is the value of the derivative at that point.

Key Takeaways

The derivative of $\ln x$ is $\frac{1}{x}$. For a composite function $\ln(ax+b)$, the chain rule gives $\frac{a}{ax+b}$. The gradient of a curve at a given $x$-value is found by substituting that value into the derivative.

Common Mistakes

Forgetting the chain rule and writing the derivative as $\frac{1}{5x+1}$ instead of $\frac{5}{5x+1}$. Substituting $x = 4$ into the original function rather than into the derivative. Making an arithmetic slip in the denominator: $5(4)+1 = 21$, not $25$ or $9$.

Things to Be Careful About

The derivative of $\ln(5x+1)$ is $\frac{5}{5x+1}$, not $\frac{1}{5x+1}$; the factor $5$ from differentiating $5x+1$ must be included. The mark scheme allows an initial derivative of the form $\frac{k}{5x+1}$ for the method mark, but the final correct derivative must be $\frac{5}{5x+1}$. Substitute $x = 4$ into the derivative, not into $y$. At $x = 4$ the argument $5x+1 = 21$ is positive, so the logarithm and its derivative are defined.