Finding F'(x) for Square Root of x

  • #1

Homework Statement


F(x)= square root of x

what would F'(x)=?
(F prime of x)


Homework Equations



F(x)= 3x^2
F'(x)=6x

F(x)= x^3
F'(x)=3x^2

The Attempt at a Solution



no clue. If the variable is a square root, it wouldn't have an exponent, right?
confused.
 
Last edited:
  • #2

Homework Statement


F(x)= square root of x

what would F'(x)=?
(F prime of x)


Homework Equations



F(x)= 3x^2
F'(x)=6x

F(x)= x^3
F'(x)=3x^2
The general equation you're after is [tex]\frac{d}{dx}(x^n)=nx^{n-1}[/tex]

The Attempt at a Solution



no clue. If the variable is a square root, it wouldn't have an exponent, right?
confused.

Have you not come across fractional powers? i.e. the nth root of x is denoted x1/n?
 
  • #3
I have, just failed to make the connection. Calculus is not my strong point. Thank you very much for your help.
 

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