# Thread: Differentiate

1. ## Differentiate

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

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

a)

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

But the book's answer is $\frac{1}{2x}$ .
Is my answer wrong?

b)

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

But the book's answer is $-\frac{1}{\sqrt 2}$ .
Is my answer wrong again?

4. Originally Posted by SengNee
a)

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

But the book's answer is $\frac{1}{2x}$ .
Is my answer wrong?

b)

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

But the book's answer is $-\frac{1}{\sqrt 2}$ .
Is my answer wrong again?
I agree with you for the answer to both problems.

-Dan

5. Originally Posted by topsquark
I agree with you for the answer to both problems.

-Dan
Seconded.