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>
Since $\displaystyle \log{x^n} = n\log{x}$, for all $\displaystyle n,x\in\mathbb{R}_{>0}$, we have $\displaystyle f(x) = \frac{1}{2}\log(1-x)$.1. If $\displaystyle f(x) = \log\sqrt{1-x}$, find $\displaystyle f'(x)$.
Then using the chain-rule we find $\displaystyle f'(x) = -\frac{1}{2(1-x)}$.
Write it as $\displaystyle f(x) = k^2\{\sin(2x)\}^2+bx$. Then (again) by the chain-rule,2. Let $\displaystyle f(x) = k^2 sin^2(2x) + bx$, where $\displaystyle x\in\mathbb{R}$ and k, b are non-zero constants.
$\displaystyle f'(x) = (2k^2\sin{2x})(2\cos{2x})+b = 4k^2\sin{2x}\cos{2x}+b = 2k^2\sin{4x}+b. $
Note that it's equavalent to $\displaystyle \frac{1}{2}\sin{4x}$ - hence $\displaystyle f'(x) = 2\cos{4x}$ and $\displaystyle f''(x) = -8\sin{4x}$. Then3. For $\displaystyle x\in\mathbb{R}$, find the minimum value of $\displaystyle f(x) = \sin{2x}\cos{2x}$.
we have $\displaystyle f'(x) = 0 \Rightarrow x = \frac{n\pi}{4}-\frac{\pi}{8}, n\in\mathbb{Z}$. Recall that a minimum of $\displaystyle f(x)$ is a point $\displaystyle b$ such that
$\displaystyle f'(b) = 0$ and $\displaystyle f''(b) > 0$. Take the first zero of $\displaystyle f'(x)$, that is $\displaystyle x = -\frac{\pi}{8}$, and we have $\displaystyle f''(x) = 8$,
so it is a minimum. Thus the minimum value of $\displaystyle f(x)$ is $\displaystyle \frac{1}{2}\sin\left(\frac{-4\pi}{8}\right) = -\frac{1}{2}$.
By the chain-rule (again) the derivative of [tex]e^{\frac{x}{3}}[/Math] is $\displaystyle \frac{1}{3}e^{\frac{x}{3}}$ (fill the details). Since $\displaystyle \log{x^n} = n\log{x}$ for any4. Find $\displaystyle f'(x)$ if $\displaystyle f(x) = e^{\frac{x}{3}}-2\log\left(\sqrt{\frac{x}{3}}\right)$
positive real $\displaystyle n$, we have $\displaystyle -2\log\left(\sqrt{\frac{x}{3}}\right) = \log\left(\frac{3}{x}\right) = \log{3}-\log{x}$ and the derivative of that is
(obviously) $\displaystyle -\frac{1}{x}$. Thus $\displaystyle f'(x) = \frac{1}{3}e^{\frac{x}{3}}-\frac{1}{x}$. (It's late over here so I apologise if there happen to be any
'mistypes' or mistakes).