The function \(y\) is defined by \(y=3x-2x^2\). Write down its derivative and show that, when \(x=5\), a small increase in \(x\) of \(p\%\) causes an increase in the magnitude of \(y\) of approximately \(17p/7\%\).

Suppose \(y = f(x)\), and that \((x,y)\) is on this curve. We now increase \(x\) by a small amount \(\delta x\), which increases \(y\) by a small amount \(\delta y\) (\(y\) could decrease, in which case \(\delta y\) would be negative).

Can you see why \(y'\) evaluated at \(x\) is roughly \(\dfrac{\delta y}{\delta x}\)? A diagram might help…