Hi guys,
I need some help with a past hsc question. It's from 2008 (Q10-b-ii) for math advance. My problem is with the differentiation part, when we have to differentiate the area A. I looked through the answers and the marking centre feedback but I still don't get it.
If S, L and Sin are constants, why they are not cancelled in the derivative?
And why they cancelled one L inside the brackets but left the other one squared?
I hope my question makes sense. If anyone can help me with this, it would be much appreciated. Thanks.
I'm afraid I'm not 100% certain about the issue, but it sounds like the differentiation itself is what you're asking about?
\[ \text{As }s \text{ and }\sin \alpha\text{ are constants, we can move them outside the derivative}\\ \text{using the rule }\frac{d}{dx}(cf(x)) = c\frac{d}{dx}f(x) \]
\begin{align*}
\frac{dA}{dx} &= \frac{d}{dx} \left[s\left(x-\ell +\frac{\ell^2}{2x} \right)\sin\alpha\right]\\
&= s\sin \alpha \times \frac{d}{dx}\left(x-\ell + \frac{\ell^2}{2x} \right)
\end{align*}
\[ \text{Now recall that the derivative is with respect to }x.\\ \text{As always, when there's a sum of terms, we differentiate term by term.}\\ \text{Firstly, }\frac{d}{dx} x = 1. \]
\[ \text{Then, }\frac{d}{dx}\ell = 0\text{, because }\ell\text{ is a constant.}\\ \text{So differentiating }\ell\text{ has the same effect as differentiating a number, like }2. \]
\[ \text{With the last term, the }\frac{\ell^2}{2}\text{ bit is also a constant.}\\ \text{So we move it out in front.}\\ \frac{d}{dx} \left( \frac{\ell^2}{2x} \right) = \frac{\ell^2}{2} \times \frac{d}{dx} \frac1x \]
\[ \text{And since }\frac1x\text{ is all we're left with, we just use the power rule to deduce that}\\ \frac{d}{dx} \frac1x = -\frac1{x^2}. \]
Which gives the final answer \( \frac{dA}{dx} = s\sin \alpha \times \left(1-\frac{\ell^2}{2x^2} \right) \)
Make sure you understand that the derivative is with respect to \(x\). Not with respect to something else like \(\ell\). The \(\ell\) inside was not cancelled because of the fact it was an \(\ell\), but rather because
it was a constant when as opposed to \(x\) and hence its derivative is
zero. Whereas that \( \frac{\ell^2}{2x}\) term actually has an \(x\) attached to it, and hence requires use of laws of derivatives.