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