1. ## Differentiate

Differentiate:
a) $\displaystyle \ln\left({\frac{2}{\sqrt{x}}}\right)$
b) $\displaystyle \sin x-2\cos x$ , $\displaystyle x=\frac{\pi}{4}$

2. Expand the first one before takin' its derivative. As for the second one it's just a matter of basic derivatives, check it out in your notes.

3. Differentiate:
a) $\displaystyle \ln\left({\frac{2}{\sqrt{x}}}\right)$
b) $\displaystyle \sin x-2\cos x$ , $\displaystyle x=\frac{\pi}{4}$

a)

$\displaystyle \frac{d}{dx}\left[\ln\left({\frac{2}{\sqrt{x}}}\right)\right]$
$\displaystyle =\frac{-\frac{1}{2}(2x^{-\frac{3}{2}})}{2x^{-\frac{1}{2}}}$
$\displaystyle =-\frac{1}{2x}$

But the book's answer is $\displaystyle \frac{1}{2x}$ .

b)

$\displaystyle \frac{d}{dx}\left(\sin x-2\cos x\right)$
$\displaystyle =\cos x +2\sin x$
$\displaystyle =\frac{1}{\sqrt 2}+2\left(\frac{1}{\sqrt 2}\right)$
$\displaystyle =\frac{3}{\sqrt 2}$

But the book's answer is $\displaystyle -\frac{1}{\sqrt 2}$ .

4. Originally Posted by SengNee
a)

$\displaystyle \frac{d}{dx}\left[\ln\left({\frac{2}{\sqrt{x}}}\right)\right]$
$\displaystyle =\frac{-\frac{1}{2}(2x^{-\frac{3}{2}})}{2x^{-\frac{1}{2}}}$
$\displaystyle =-\frac{1}{2x}$

But the book's answer is $\displaystyle \frac{1}{2x}$ .

b)

$\displaystyle \frac{d}{dx}\left(\sin x-2\cos x\right)$
$\displaystyle =\cos x +2\sin x$
$\displaystyle =\frac{1}{\sqrt 2}+2\left(\frac{1}{\sqrt 2}\right)$
$\displaystyle =\frac{3}{\sqrt 2}$

But the book's answer is $\displaystyle -\frac{1}{\sqrt 2}$ .