Small Changes and Approximation

The derivative $\frac{dy}{dx}$ is a rate of change, so for a small step $\delta x$ in $x$, the change in $y$ is approximately $\delta y \approx \frac{dy}{dx}\,\delta x$. This is the small-changes technique: it treats the curve as its tangent line over a short interval.

Given $y = x^3 + 2x$, use calculus to find the approximate change in $y$ as $x$ increases from $3$ to $3.05$. [4]

The working

Step 1, differentiate: $\frac{dy}{dx} = 3x^2 + 2 \quad \text{(M1)}$

Step 2, evaluate at the starting value $x = 3$: $\left.\frac{dy}{dx}\right|_{x=3} = 3(9) + 2 = 29 \quad \text{(A1)}$

Step 3, identify $\delta x$: the step is $3.05 - 3 = 0.05$. (B1)

Step 4, apply $\delta y \approx \frac{dy}{dx}\,\delta x$: $\delta y \approx 29 \times 0.05 = 1.45 \quad \text{(A1)}$

So $y$ increases by approximately $1.45$.

Where the marks are won and lost

• Evaluate the derivative at the starting $x$ ($x = 3$), not the end value.
• $\delta x$ is the change in $x$ ($0.05$), not the new value of $x$.
• The answer is a change in $y$, not a new value of $y$. If asked for the new $y$, add $\delta y$ to the original.

Common mistakes

• Using $\delta x = 3.05$ instead of $0.05$.
• Multiplying by $x$ instead of by $\frac{dy}{dx}$.
• Trying to compute the exact change (this question asks for the calculus approximation).

Topic home: Calculus pillar. More: Worked examples.