Derivatives?

Question

find f1(x)

1. f(x) = (x^3 -2x+1)^4
2. f(x) = x^2 (x + 1)^-1
3. f(x) = (x^2 - 4) ^-1/2
4. f(x) = x+1 / x^2 - 4
5. f(x) = (x+1 / x-1)
6. f(x) = x sqrt(1-x^3)

Answer 1

if you mean find f '(x) than my answer well be
1. f(x) = (x^3 -2x+1)^4
f '(x) = 4(x^3 -2x+1)^3 * 3x^2 - 2 + 0
2. f(x) = x^2 (x + 1)^-1
f ' (x) = 2x * (x+1)^-1 + x^2 * -(x+1)^-2
3. f(x) = (x^2 - 4) ^-1/2
f ' (x) = -1/2(x^2 - 4)^-3/2
4. f(x) = x+1 / x^2 - 4
f ' (x) = [1 * x^2 - 4 - (x+1) * (2x)] / (x^2-4)^2
5. f(x) = (x+1 / x-1)
f ' (x) = [1 * (x-1) - (x+1) * (1)] / (x-1)^2
6. f(x) = x sqrt(1-x^3)
f(x) = x(1-x^3)^1/2
f ' (x) = 1 * (1-x^3)^1/2 +x * 1/2(1-x^3)^3/2 * 3x^2
did you ask this question only to check your answer
or you do not how to do it
Because if you do not know than ask for basic equations of these derivatives
Please you need them a lot when you start dealing with integrals

Answer 2

1. 4(3x^2-2)(x^3-2x+1)^3

The others work the same way. Remember that the square root is the same thing as to the 1/2 power.

Answer 3

These all call for combinations of the product rule, quotient rule, and chain rule. They are as follows.

Product Rule: f(x) = g(x)*h(x) ==> f'(x) = g(x)*h'(x) + g'(x)*h(x)
Quotient Rule: f(x) = g(x) / h(x) ==> f'(x) = [h(x)*g'(x) - g(x)*h'(x)] / (h(x))^2
Chain Rule: f(x) = g(h(x)) ==> f'(x) = g'(h(x))*h'(x)

For example, 1. requires the chain rule. In this case, g(x) = x^4 ==> g'(x) = 4x^3 and h(x) = x^3 - 2x + 1 ==> h'(x) = 3x^2 - 2. So f'(x) = g'(h(x))*h'(x) = 4[(x^3 - 2x + 1)^3](3x^2 - 2).

2. can be done either with the product and chain rules, or with only the quotient rule if you rewrite (x + 1)^-1 as 1 / (x + 1). 3. is chain rule. 4. is quotient rule. 5. is quotient rule. 6. is product and chain rules.

Answer 4

Hi,
1. f'(x)=4·(x^3 -2x+1)^3·(3x^2-2)
2. f'(x)=(2x·(x+1)-x^2)/(x+1)^2
f'(x)=(x^2+2x)/(x+1)^2
3. f'(x)=-1/2·(x^2 - 4)^-3/2·(2x)
f'(x)=-x·(x^2 - 4)^-3/2
4.f`(x)=1-2/x^3
5.f'(x)=((x-1)-(x+1))/(x-1)^2
f'(x)=-2/(x+1)^2
6.f'(x)=sqrt(1-x^3)+x·[(1/2·(-3x^2))/sqrt(1-x^3)]
f'(x)=[(1-x^3)+(x/2-3x^3)]/sqrt(1-x^3)
f'(x)=[1+x/2-4x^3]/sqrt(1-x^3)

good luck