The two variables $x$ and $y$ are such that $y = \frac{10}{(x + 4)^3}$. (i) Find an expression for $\frac{\mathrm{d}y}{\mathrm{d}x}$. (ii) Hence find the approximate change in $y$ as $x$ increases from $6$ to $6 + p$, where $p$ is small.

(i) dy/dx = -30/(x + 4)^4 · (ii) -0.003p

Additional Mathematics 4037/23 — October/November 2010

Cambridge O-Level · Calculator · worked solutions for every part, with the mark scheme

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Topics Calculus · Straight-line graphs · Trigonometry · Series · [Legacy] Matrices · [Legacy] Set language and notation · +5 more

Q1CalculusFree sample

The two variables $x$ and $y$ are such that $y = \frac{10}{(x + 4)^3}$ .

(i)

Find an expression for $\frac{\mathrm{d}y}{\mathrm{d}x}$ .

DifficultyMedium-Easy

Worked solution

Approach

Rewrite $y$ as a negative power and differentiate $(x+4)^{-3}$ using the chain rule.

Working

y = \frac{10}{(x+4)^3} = 10(x+4)^{-3}

Differentiating:

\frac{\mathrm{d}y}{\mathrm{d}x} = 10 \times (-3)(x+4)^{-4} = -30(x+4)^{-4}

Answer

\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{30}{(x+4)^{4}}

Final answer

dy/dx = -30/(x + 4)^4

Detailed explanation

Walkthrough

The function is given as a fraction, so the first step is to rewrite it in index form: dividing by $(x+4)^3$ is the same as multiplying by $(x+4)^{-3}$ . This puts $y$ into the form $k(x+4)^n$ , which differentiates cleanly. Applying the chain rule — multiply by the power, reduce the power by one, then multiply by the derivative of the inner bracket (which is just $1$ ) — gives $10 \times (-3)(x+4)^{-4} = -30(x+4)^{-4}$ , i.e. $-\frac{30}{(x+4)^4}$ .

Key Takeaways

Convert reciprocal powers to negative indices before differentiating.
The chain rule on $(ax+b)^n$ : bring down $n$ , reduce the power, multiply by $a$ .

Common Mistakes

Forgetting the factor of $-3$ from the power, or dropping the minus sign.
Leaving the answer as $(x+4)^{-4}$ without the coefficient $-30$ .
Differentiating only the numerator and ignoring the denominator structure instead of rewriting as an index first.

Things to Be Careful About

The mark scheme expects the form $k(x+4)^{-4}$ with $k = -30$ ; keep the exact coefficient rather than a decimal.
Writing the final answer back over a positive power, $-\frac{30}{(x+4)^4}$ , is equivalent and acceptable.

Techniques used

rewrite the function as a power of a linear expressiondifferentiate using the chain rule

(ii)

Hence find the approximate change in $y$ as $x$ increases from $6$ to $6 + p$ , where $p$ is small.

DifficultyMedium-Easy

Worked solution

Approach

Use the small-change approximation $\delta y \approx \frac{\mathrm{d}y}{\mathrm{d}x}\,\delta x$ , evaluating the derivative from part (i) at $x = 6$ with $\delta x = p$ .

Working

Using the result from part (i), $\frac{\mathrm{d}y}{\mathrm{d}x} = -30(x+4)^{-4}$ :

At $x = 6$ :

\frac{\mathrm{d}y}{\mathrm{d}x} = -30(6+4)^{-4} = -30 \times 10^{-4} = -\frac{30}{10000}

With $\delta x = p$ :

\delta y \approx -\frac{30}{10000} \times p = -0.003p

Answer

\delta y \approx -0.003p

Final answer

-0.003p

Detailed explanation

Walkthrough

The word "Hence" tells us to use the derivative found in part (i). Since $p$ is small, the change in $y$ can be approximated by $\delta y \approx \frac{\mathrm{d}y}{\mathrm{d}x}\,\delta x$ — the derivative measures the local rate of change, so multiplying it by the small increase in $x$ estimates the resulting increase in $y$ . Here $\delta x = p$ and we evaluate the gradient at the starting value $x = 6$ : substituting gives $-30(10)^{-4} = -0.003$ . Multiplying by $p$ gives the approximate change $-0.003p$ ; the negative sign shows that $y$ decreases as $x$ increases.

Key Takeaways

The small-change formula: $\delta y \approx \frac{\mathrm{d}y}{\mathrm{d}x}\,\delta x$ .
Evaluate the derivative at the starting value of $x$ , not the new value.
A negative result means $y$ decreases when $x$ increases.

Common Mistakes

Evaluating the derivative at $x = 6 + p$ instead of $x = 6$ .
Arithmetic slips with the power of ten: $10^{-4} = 0.0001$ , so $-30 \times 0.0001 = -0.003$ .
Dropping the negative sign, which loses the accuracy mark.
Treating $p$ as a number to substitute rather than leaving the answer in terms of $p$ .

Things to Be Careful About

The mark scheme awards the M1 for correctly setting up $\delta y = \frac{\mathrm{d}y}{\mathrm{d}x} \times \delta x$ with $x = 6$ and $\delta x = p$ ; show this substitution explicitly.
The A1 is a follow-through mark (√) on the candidate's part (i) derivative, but the printed answers are $-0.003p$ or $-\frac{3}{1000}p$ ; give the answer to at least 3 significant figures if decimal.

Techniques used

apply the small-increment approximationevaluate the derivative at the given pointexpress the change in terms of the parameter p

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