Homework 4 was due on Thursday 19th September 2013 and consisted of:
I marked 3.3/34 out of 3, 3.3/48 out of 2, 3.5/28 out of 3, and 3.5/52 out of 2.
Section 3.3 Q34. Most people calculated $f'(x) = e^x(\cos x - \sin x)$ correctly using the product rule; this made me happy. What made me sad is that most people dropped a point on the last part. You needed to find all the values of $x$ for which $e^x(\cos x - \sin x) = 0$. Since $e^x \ne 0$ for any value of $x$, we can freely divide by it. Rearranging what remains gives $\sin x = \cos x$. Most people got this far without any problems. We can simplify this equation to give $\tan x = 1$. Now $x=\frac{\pi}{4}$ is a solution to this equation; then, using the fact that $\tan x$ is periodic with period $\pi$, we get that the solution to the question is $$x = \frac{\pi}{4} + n \pi \quad \text{for integer values of}\ n$$ Many people either didn't use periodicity at all, or used the wrong period (say $\frac{\pi}{2}$ or $2\pi$), and thus either provided values of $x$ for which $f'(x) \ne 0$, or didn't provide some solutions at all.
Section 3.3 Q48.The intended method for this question was to use the result that $$\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$$ Putting $x-1 = \theta$ we see that $x \to 1$ as $\theta \to 0$, so we get the result $$\lim_{x \to 1} \frac{\sin (x-1)}{x-1} = 1$$ This then leads to the desired result. Quite a few people tried to use L'Hôpital's rule. This gives the correct answer, but I was told by the instructor to deduct a point for doing this. The reason for this is that L'Hôpital's rule implicitly uses the result you're trying to prove: in order to compute the derivative of $\sin(x-1)$ from the definition, you use the result that $\lim_{x \to 1} \frac{\sin (x-1)}{x-1} = 1$. As such, using L'Hôpital's rule here is circular logic.
Section 3.5 Q28. This question was done very well on the whole. The main errors were either not using the product rule on $2xy$ properly, or making small arithmetical mistakes... a lot of people didn't keep track of their $+$ and $-$ signs very well, for instance!
Section 3.5 Q52. The answer you get for this question depends on which variant of the function $\sec^{-1}(x)$ you use: see Section 3.5 Q64 for more details. I accepted both variants, but in the end it didn't make much difference. Most mistakes on this question were small algebraic errors or misapplications of the chain or product rules.