# Homework Help: Find Derivatives of a Function

1. Oct 20, 2011

### Joyci116

1. The problem statement, all variables and given/known data
If f(x)=$\sqrt{x}$g(x), where g(4)=8 and g'(4)=7, find f'(4).

2. Relevant equations
the product rule/quotient rule

3. The attempt at a solution
I am having trouble getting this problem started.
Would I solve for the x of f(x)?

2. Oct 20, 2011

### gb7nash

Forget the numbers for a second. Let's look for f'(x). What does the product rule say?

3. Oct 20, 2011

### Joyci116

The product rule is
d/dx [f(x)*g(x)]=f(x)g'(x)+g(x)f'(x)

4. Oct 20, 2011

### gb7nash

Ok, good. Let's apply this to the problem now.

5. Oct 20, 2011

### Joyci116

f(x)=x$^{\frac{1}{2}}$g(x)
f'(x)=$\frac{1}{2}$x$^{-\frac{1}{2}}$g(x)

6. Oct 20, 2011

### gb7nash

Just so you don't get confused with using the same letters, we'll say the product rule is:

d/dx[p(x)q(x)] = p(x)q'(x) + q(x)p'(x)

This is only one part of it. Let p(x) = x1/2 and q(x) = g(x). What are you missing?

7. Oct 20, 2011

### Joyci116

f'(x)=x$^{\frac{1}{2}}$7+8($\frac{1}{2}$x$^{-\frac{1}{2}}$)

8. Oct 20, 2011

### gb7nash

You're partially skipping a step. The 7 and the 8 are what you get when you want to know what f'(4) is. We're just looking for the general derivative with respect to x. What's f'(x) now?

9. Oct 20, 2011

### Joyci116

f'(x)=X^(1/2)g'g(x)+g(x)(1/2)x^(-1/2) ?

10. Oct 20, 2011

### gb7nash

Fixed a typo you made. That looks fine. So what's f'(4)?

11. Oct 20, 2011

### Joyci116

f'(x)=X^(1/2)g'(x)+g(x)(1/2)x^(-1/2) ?

12. Oct 20, 2011

### Joyci116

16!

Thank you very much :)