# Differentiate

• Mar 23rd 2008, 05:13 PM
SengNee
Differentiate
Differentiate:
a) $\ln\left({\frac{2}{\sqrt{x}}}\right)$
b) $\sin x-2\cos x$ , $x=\frac{\pi}{4}$
• Mar 23rd 2008, 05:16 PM
Krizalid
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.
• Mar 23rd 2008, 05:53 PM
SengNee
Quote:

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 $\frac{1}{2x}$ .

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 $-\frac{1}{\sqrt 2}$ .
• Mar 23rd 2008, 06:05 PM
topsquark
Quote:

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 $\frac{1}{2x}$ .

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 $-\frac{1}{\sqrt 2}$ .

I agree with you for the answer to both problems.

-Dan
• Mar 23rd 2008, 06:28 PM
mr fantastic
Quote:

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

-Dan

Seconded.