Differentiation

Archimedes · 18-08-2009, 15:04

Guys, my differentiation is rusty. What is the differentiation of

a) x/7

b) 1/4 √x

**-JammyDodger-** · 18-08-2009, 15:13

Quote:

Originally Posted by Archimedes

a) x/7

Well, $\displaystyle\frac{x}{7}$ is the same thing as $\displaystyle\frac{1}{7}\mbox{x}$ . So, differentiate $\displaystyle\mbox{x}$ and, at the very end, multiply it by $\displaystyle\frac{1}{7}$ . So, what's the derivative of x with respect to x?

Quote:

b) 1/4 √x

For this one, change the $\displaystyle\sqrt{x}$ into its indice form, i.e. $x^{\frac{1}{2}}$ . Now, differentiate $x^{\frac{1}{2}}$ and, again at the very end, multiply it by $\displaystyle\frac{1}{4}$ .

When there's a constant (a number) in front of the variable (in both of these cases, x), you can, basically, ignore it until the end and then multiply it by the derivative of the x.

Sean_K · 18-08-2009, 15:16

if $\text{[math]}$ then $f'(x) = n x^{n-1}$ .
and
$\text{[math]}$
$\sqrt{x} = x^{1/2}$
/edit: beaten to it

gerry87 · 18-08-2009, 15:20

y= x/7
dy/dx = 1/7

for the 2nd one is sqrt(x) above or below the line?

y = (1/4) * x^.5
dy/dx = .5*(1/4) * x^(-.5)
dy/dx = 1/(8*sqrt(x))

You can do sums like this in wolfram alpha http://www.wolframalpha.com/input/?i...iate+y%3Dx%2F7

Archimedes · 18-08-2009, 15:45

I send many thanks and best wishes in all your respective directions

18-08-2009, 15:04	#1
Archimedes Registered User Join Date: Jun 2006 Location: My heart is in Kilkenny but my stuff is in Belfast Posts: 4,884 Adverts \| Friends	Differentiation Guys, my differentiation is rusty. What is the differentiation of a) x/7 b) 1/4 √x __________________

18-08-2009, 15:16	#3
Sean_K Registered User Join Date: Jul 2005 Location: Co. Dublin Posts: 5,524 Adverts \| Friends	if $\text{[math]}$ then $f'(x) = n x^{n-1}$ . and $\text{[math]}$ $\sqrt{x} = x^{1/2}$ /edit: beaten to it __________________ *Dropbox: Free file backup, sync and storage* This is a referral link. Sign up using it and we both get an extra 0.25 GiB

18-08-2009, 15:45	#5
Archimedes Registered User Join Date: Jun 2006 Location: My heart is in Kilkenny but my stuff is in Belfast Posts: 4,884 Adverts \| Friends	I send many thanks and best wishes in all your respective directions __________________

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18-08-2009, 15:20	#4
gerry87 Registered User Join Date: Nov 2005 Posts: 829 Adverts \| Friends	y= x/7 dy/dx = 1/7 for the 2nd one is sqrt(x) above or below the line? y = (1/4) * x^.5 dy/dx = .5(1/4) x^(-.5) dy/dx = 1/(8*sqrt(x)) You can do sums like this in wolfram alpha http://www.wolframalpha.com/input/?i...iate+y%3Dx%2F7