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**1. This antiderivative question asks: Find f.**

The equation reads: f prime of x = square root of x to the 5th power minus 4 divided by the fifth root of x. (see below)

f´ (x) = √(x^5 ) - 4/√(5&x)

The answer in the back of the book is f(x) = (2/7)X^(7/2) - 5X^(4/5) + C but I got stuck trying to work it out.

The equation reads: f prime of x = square root of x to the 5th power minus 4 divided by the fifth root of x. (see below)

f´ (x) = √(x^5 ) - 4/√(5&x)

The answer in the back of the book is f(x) = (2/7)X^(7/2) - 5X^(4/5) + C but I got stuck trying to work it out.

2. Here are the relevant equations:

If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is: F(x) + C

**Antidifferentiation formulars:**

Function = cf(x) Particular antiderivative = cF(x)

Function = f(x) + g(x) Particular antiderivative = F(x) + G(x)

Function = x^n (n not equal -1) Particular antiderivative (x^n+1)/n+1

Function = cf(x) Particular antiderivative = cF(x)

Function = f(x) + g(x) Particular antiderivative = F(x) + G(x)

Function = x^n (n not equal -1) Particular antiderivative (x^n+1)/n+1

3. My attempt at a solution reads as follows:

**X to the fifth power raised to the half power minus 4 times 5X raised to the minus a half**

= 〖〖(X〗^(5))〗^(1/(2 ))-4(5X^((-1)/2))

= 〖〖(X〗^(5))〗^(1/(2 ))-4(5X^((-1)/2))

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