Math final help.?

Question

1. find the derivative of f '(x)
exaplin. x^2 +20, x=3

Answer 1

ok, well, you need to take the derivative, here you use the exponent rule...that means you take the exponent, bring it down front, and decrease by one, the derivative of a constant (a number) is zero...so deriv. of 20 is 0...so

x^n=n*x^(n-1) ,--exponent rule

so, x^2=2*x^(2-1) => 2*x

and deriv. of 20 is 0, so your derivative is 2x+0, now you want to evaluate at 3, so plug 3 in for x, you get

2(3)+0=6

hope this helps!

Answer 2

f(x)=x^2 +20
f '(x)=2x

derivative rule for x^n where n is any integer is f'(x)=n*x^n-1
when you have f(x)=g(x)+h(x) f'(x)= g'(x)+h'(x) as is if there is an addition of two terms you can find the derivative of each one and then add them together.

so in the case x=3, f '(x)=2x f'(3)=2(3)=6

Answer 3

f(x) = x^2 +20
take the derivative, so
f'(x) = 2x
now substitute for x
f'(3) = 2*3 = 6

Answer 4

2x, for x=3, the answer would then be 6