Mathematics 9709/32 — October/November 2013

Walkthrough

We are asked to show that the gradient, i.e. the derivative $\frac{dy}{dx}$, is negative for every $x > -\frac{1}{2}$. The function is a quotient of two linear expressions, so the quotient rule is the natural tool.

First identify the numerator and denominator: $u = 1+x$, $v = 1+2x$. Differentiate each: $u' = 1$, $v' = 2$.

Apply the quotient rule:

$$\frac{dy}{dx} = \frac{v u' - u v'}{v^2}$$

Substitute and simplify:

$$\frac{dy}{dx} = \frac{(1+2x) - 2(1+x)}{(1+2x)^2} = \frac{-1}{(1+2x)^2}$$

Now examine the sign. For any real $x$ except $x = -\frac12$, the square $(1+2x)^2$ is positive. Since the domain is $x > -\frac12$, $1+2x \neq 0$, so the denominator is positive. The numerator is $-1$, which is negative. Therefore the whole derivative is negative.

This matches the three marks: M1 for using the quotient/product rule, A1 for obtaining the correct derivative, and A1 for justifying the sign.

Key Takeaways

• A quotient of functions is differentiated with the quotient rule.
• The sign of a derivative of the form $\frac{-1}{(1+2x)^2}$ depends only on the sign of the numerator because the squared denominator is always non-negative when nonzero.
• Domain restrictions matter: here $x > -\frac12$ ensures the denominator is never zero.

Common Mistakes

• Using the quotient rule in the wrong order, e.g. writing $\frac{u'v - uv'}{v^2}$ instead of $\frac{vu' - uv'}{v^2}$.
• Forgetting to square the denominator.
• Incorrectly simplifying $1+2x - 2 - 2x$ to something other than $-1$.
• Saying the derivative is negative just because the numerator is negative, without noting that the denominator is positive.

Things to Be Careful About

• The denominator $(1+2x)^2$ is positive for all $x$ except $x = -\frac12$; the domain excludes this value, so no division by zero occurs.
• The derivative can also be found by rewriting $y$ using the product rule as $y = (1+x)(1+2x)^{-1}$; the quotient rule is usually simpler here.
• The mark scheme requires a clear justification that the derivative is negative, not just the formula.