# differential calculus

• Jan 21st 2008, 06:30 AM
izaiyoi
differential calculus
differentiate
y=(10x^4 + 40x^3)^5 + 100x
• Jan 21st 2008, 06:35 AM
Krizalid
You know what the power rule is? Chain Rule?

You have two options: differentiate as it's written (you'll need the Chain Rule) or expand and apply easily the power rule. (I prefer the first option, of course.)
• Jan 21st 2008, 06:53 AM
colby2152
Quote:

Originally Posted by Krizalid
You know what the power rule is? Chain Rule?

You have two options: differentiate as it's written (you'll need the Chain Rule) or expand and apply easily the power rule. (I prefer the first option, of course.)

You will definitely want to use the chain rule here:

$f = u^5+v$
$f'=5u^4*u'+v'$
• Jan 21st 2008, 12:32 PM
mr fantastic
Quote:

Originally Posted by Krizalid
You know what the power rule is? Chain Rule?

You have two options: differentiate as it's written (you'll need the Chain Rule) or expand and apply easily the power rule. (I prefer the first option, of course.)

The thanks is for your current signature - let's hope it proves to be most useful.
• Jan 21st 2008, 06:24 PM
memena
not sure but here i go
y=(10x^4 + 40x^3)^5 + 100x

y'= u need to use chain rule= derivative of outer times original inner times derivative of inner...= DOI x DI ...in this case

the outer function or number is 5 and the inner function is (10x^4 + 40x^3), this one is easy to tell which is outer and inner b/c the inner function is in parentheses. Remember to follow the power rule too, nx^n-1

y'= 5(10x^4 + 40X^3)^5-1 = 5(10x^4 + 40X^3)^4 ok this is the first half of chain rule, then you multiply by the derivative of the inner function using the power rule.

y'= 5(10x^4 + 40X^3)^4 * (40x^4-1 +120^3-1)

y'= 5(10x^4 + 40X^3)^4 * (40x^3 +120^2)...

ok almost done now you want find the derivative of 100x which is simply 100.

so ur final answer is :

y'= 5(10x^4 + 40X^3)^4 * (40x^3 +120^2) + 100

I hope that helps...later

God is love,
Mary