1. ## Further differentiation

Q1)

Q2) Let f(x) = k^2 sin^2(2x) + bx, where x ε R and k, b are non zero
constants
a) Find f’(x)
b) For x ε R, the minimum value of sin(2x)cos(2x) is
Q3) <attached image>

2. 1. If $f(x) = \log\sqrt{1-x}$, find $f'(x)$.
Since $\log{x^n} = n\log{x}$, for all $n,x\in\mathbb{R}_{>0}$, we have $f(x) = \frac{1}{2}\log(1-x)$.
Then using the chain-rule we find $f'(x) = -\frac{1}{2(1-x)}$.

2. Let $f(x) = k^2 sin^2(2x) + bx$, where $x\in\mathbb{R}$ and k, b are non-zero constants.
Write it as $f(x) = k^2\{\sin(2x)\}^2+bx$. Then (again) by the chain-rule,
$f'(x) = (2k^2\sin{2x})(2\cos{2x})+b = 4k^2\sin{2x}\cos{2x}+b = 2k^2\sin{4x}+b.$

3. For $x\in\mathbb{R}$, find the minimum value of $f(x) = \sin{2x}\cos{2x}$.
Note that it's equavalent to $\frac{1}{2}\sin{4x}$ - hence $f'(x) = 2\cos{4x}$ and $f''(x) = -8\sin{4x}$. Then
we have $f'(x) = 0 \Rightarrow x = \frac{n\pi}{4}-\frac{\pi}{8}, n\in\mathbb{Z}$. Recall that a minimum of $f(x)$ is a point $b$ such that
$f'(b) = 0$ and $f''(b) > 0$. Take the first zero of $f'(x)$, that is $x = -\frac{\pi}{8}$, and we have $f''(x) = 8$,
so it is a minimum. Thus the minimum value of $f(x)$ is $\frac{1}{2}\sin\left(\frac{-4\pi}{8}\right) = -\frac{1}{2}$.

4. Find $f'(x)$ if $f(x) = e^{\frac{x}{3}}-2\log\left(\sqrt{\frac{x}{3}}\right)$
By the chain-rule (again) the derivative of $$e^{\frac{x}{3}}$$ is $\frac{1}{3}e^{\frac{x}{3}}$ (fill the details). Since $\log{x^n} = n\log{x}$ for any
positive real $n$, we have $-2\log\left(\sqrt{\frac{x}{3}}\right) = \log\left(\frac{3}{x}\right) = \log{3}-\log{x}$ and the derivative of that is
(obviously) $-\frac{1}{x}$. Thus $f'(x) = \frac{1}{3}e^{\frac{x}{3}}-\frac{1}{x}$. (It's late over here so I apologise if there happen to be any
'mistypes' or mistakes).