Find Derivatives of a Function

[Joyci116]

Homework Statement

If f(x)=$\sqrt{x}$g(x), where g(4)=8 and g'(4)=7, find f'(4).

Homework Equations

the product rule/quotient rule

The Attempt at a Solution

I am having trouble getting this problem started.
Would I solve for the x of f(x)?

 

[gb7nash]

If f(x)=$\sqrt{x}$g(x)

the product rule/quotient rule

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

 

[Joyci116]

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

 

[gb7nash]

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

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

 

[Joyci116]

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

 

[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)

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

This is only one part of it. Let p(x) = x^1/2 and q(x) = g(x). What are you missing?

 

[Joyci116]

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

 

[gb7nash]

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

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?

 

[Joyci116]

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

 

[gb7nash]

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

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

 

[Joyci116]

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

 

[Joyci116]

16!

Thank you very much :)