Building blocks

## Things you may have noticed

Try applying the rule you came up with in the Warm-up to these functions.

\begin{align} y&=(2x-3)(x+2) \label{eq:notice1}\\ y&=(2-x)(x+3) \\ y&=(x+3)(1-2x) \\ y&=(3x+4)(2x-5) \\ y&=(3-2x)(5-3x) \end{align}
• Does your original rule give you the correct derivative for these functions?
• If your original rule works for these functions, can you justify why? If not, then can you adapt your rule to make it work for all of the examples so far?

From the Warm-up, we had a rule that said, add together the brackets to find the derivative.

If we start to look at these, we don’t seem to be able to use this rule. In $\eqref{eq:notice1}$, we see that the expansion will give us $2x^2$ so our derivative needs to include $4x$, but following the rule of adding the brackets would give us $3x$.

We could check it out thoroughly by expanding and differentiating: \begin{align*} y&=(2x-3)(x+2) \\ &=2x^2+x-6 \\ \implies\quad\frac{dy}{dx}&=4x+1 \\ \end{align*}

But $(2x-3)+(x+2)=3x-1$, which is not equal to the derivative. We notice that we’re out by $x+2$, which means that the derivative is actually $(2x-3)+2(x+2)$.

• How could we adapt our rule to work for these examples?
• Can we find a similar way of creating the derivatives of all these functions?
• Could we adapt our rule for any pair of linear brackets, or something more general (think about what that might be)?

Alternatively we could rewrite the original function to have brackets starting with only a single $x$, so for $\eqref{eq:notice1}$ we have $y=2(x-\frac{3}{2})(x+2)$.

• Now what would the adaptation to the rule be?

How are these two versions linked?