Applications of the Second Derivative?

[cmnalo]

I need to use SDT to find the relative extrema of the function:

f(x)= x square root (x-9)

I need help with the first derivative:

f'(x) = 1/2x (x-9)^-1/2

f'(x) =1/2x / (x-9)^1/2

Please help me to simplify this 1st derivative and correct me if I've worked it wrong.

 

[skeeter]

ever heard of the product rule?

\(\displaystyle \L f(x) = x(x-9)^{\frac{1}{2}}\)

\(\displaystyle \L f'(x) = \frac{x}{2}(x-9)^{-\frac{1}{2}} + (x-9)^{\frac{1}{2}}\)

\(\displaystyle \L f'(x) = \frac{x}{2\sqrt{x-9}} + \frac{2(x-9)}{2\sqrt{x-9}}\)

\(\displaystyle \L f'(x) = \frac{3(x-6)}{{2\sqrt{x-9}}\)

 

[cmnalo]

Where do you get the 3(x-6) from?

 

[skeeter]

ever heard of the product rule?

\(\displaystyle \L f(x) = x(x-9)^{\frac{1}{2}}\)

\(\displaystyle \L f'(x) = \frac{x}{2}(x-9)^{-\frac{1}{2}} + (x-9)^{\frac{1}{2}}\)

\(\displaystyle \L f'(x) = \frac{x}{2\sqrt{x-9}} + \frac{2(x-9)}{2\sqrt{x-9}} = \frac{x + 2x - 18}{2\sqrt{x-9}} = \frac{3x-18}{2\sqrt{x-9}} = ...\)

\(\displaystyle \L f'(x) = \frac{3(x-6)}{{2\sqrt{x-9}}\)

 

[tkhunny]

Where do you get the 3(x-6) from?

*** Friendly Remark Alert ***

Look in the mirror and say, "Never do that again!"

You are in calculus. Please don't get tripped up on the algebra. :wink:

 

[cmnalo]

Skeeter- Thanks for the help in clarifing how you simplified. I

tkhunny- I'm not quite sure what "Friendly Remark Alert" is? And I don't quite know what to make of your post.

 

[stapel]

tkhunny- I'm not quite sure what "Friendly Remark Alert" is?

"Warning: I'm about to say something that could be taken the wrong way. But I'm saying this to be helpful, so please take in the spirit it is meant."

Then the tutor admonished you to be mindful of what you learned back in algebra, because you still need that stuff. A lot.

Hope that helps a bit.

Eliz.

 

[tkhunny]

Wow. I meant all that? Excellent translation.