Find the derivative of f(x)=1/3 sqrt x.

The 3 is little above the square root.

I know that it is -1/3x^-4/3. Can this be written as -1/3x^4/3 ?

So I read this as

f(x) = x^(1/3)

f '(x) = (1/3)x^(-4/3)
which could be written as 1/(3x^(4/3)

why do you have a negative sign in front?

I had the negative sign there because thatg was one of the answer choices that best fit what I had but there is no negative correct? Its positive?

To find the derivative of f(x) = (1/3)√x, you can follow these steps:

1. Rewrite the function using fractional exponents:
f(x) = (1/3)x^(1/2)

2. Use the power rule to differentiate each term:
For any function g(x) = x^n, the derivative is given by g'(x) = nx^(n-1).

Applying the power rule to f(x), we get:
f'(x) = (1/3)(1/2)x^((1/2)-1)

3. Simplify the exponent:
f'(x) = (1/3)(1/2)x^(-1/2)

4. Rewrite the negative exponent by moving it to the denominator:
f'(x) = (1/3)(1/2)(1/x^(1/2))

Simplifying the expression further:
f'(x) = 1/(6√x)

Therefore, the derivative of f(x) = (1/3)√x is f'(x) = 1/(6√x).

Regarding your second question, -1/3x^(-4/3) cannot be written as -1/3x^(4/3). They have different values. To understand why, let's simplify both expressions:

-1/3x^(-4/3) can be written as (-1/3) / (x^(4/3)):
Since the exponent -4/3 is negative, we can rewrite it as the reciprocal:
-1/3 / (x^(4/3))

On the other hand, -1/3x^(4/3) is already in simplified form, as it is simply the product of -1/3 and x^(4/3).

As you can see, these two expressions have different structures and represent different values.