differentiate

y=(10x^4 + 40x^3)^5 + 100x

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- Jan 21st 2008, 07:30 AMizaiyoidifferential calculus
differentiate

y=(10x^4 + 40x^3)^5 + 100x - Jan 21st 2008, 07:35 AMKrizalid
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, 07:53 AMcolby2152
- Jan 21st 2008, 01:32 PMmr fantastic
- Jan 21st 2008, 07:24 PMmemenanot 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