Differentiation: Question 5

Worked solution

Step 1: Find $\dfrac{dy}{dx}$

$y = x^3 - 4x \implies \frac{dy}{dx} = 3x^2 - 4$

Step 2: Evaluate the derivative at $x = 3$

$\frac{dy}{dx}\bigg|_{x=3} = 3(3)^2 - 4 = 3(9) - 4 = 27 - 4 = 23$

Step 3: Apply the small increments approximation

For a small change $\delta x$, the corresponding small change in $y$ is approximated by

$\delta y \approx \frac{dy}{dx} \times \delta x$

Substituting the values found above:

$\delta y \approx 23 \times 0.02 = 0.46$

Step 4: Recompute independently as a check

Differentiating $y=x^3-4x$ from scratch: $\dfrac{d}{dx}(x^3) = 3x^2$ and $\dfrac{d}{dx}(-4x) = -4$, so $\dfrac{dy}{dx}=3x^2-4$, confirming Step 1. At $x=3$: $3\times 9 = 27$, and $27-4=23$, confirming Step 2. Then $23 \times 0.02 = 0.46$, confirming Step 3.

Independent check using the exact function values. $y(3) = 3^3 - 4(3) = 27 - 12 = 15$. For $x = 3.02$:

$3.02^3 = 27.543608, \qquad 4(3.02) = 12.08$

$y(3.02) = 27.543608 - 12.08 = 15.463608$

$\delta y_{\text{exact}} = 15.463608 - 15 = 0.463608$

This is very close to the linear approximation of $0.46$ found above. The tiny difference is due to the curvature of the function over the (small but finite) increment, which the linear approximation does not capture. This confirms $\delta y \approx 0.46$ is correct.

Why the other options are wrong

• A ($0.54$): comes from treating $-4x$ as a constant, so its derivative is taken as $0$ instead of $-4$, giving $\dfrac{dy}{dx}=27$ instead of $23$.
• C ($0.62$): comes from a sign error, differentiating $-4x$ as $+4$, giving $\dfrac{dy}{dx}=31$ instead of $23$.
• D ($0.10$): comes from forgetting to bring down the power $3$ when differentiating $x^3$, giving $\dfrac{dy}{dx}=5$ instead of $23$.

Final answer

$\boxed{\delta y \approx 0.46} \quad \text{(Option B)}$