Derivative help: derivative of -3 sqrt[x], using the def.

[Mathamateur]

Hey, I am running through some examples trying to get ready for my test and am completely stuck on one. I have a feeling something similar will be on our test too!

Anyways here is the question:

Using the definition of the derivative find f'(x). Then find f'(-2), f'(0), and f'(3).

Problem: f(x) = -3 square root of x

Sorry, I can't figure out how to make the "square root" symbol, but you get the idea: the -3 is on the outside with x inside the square root symbol. I mainly need to know f'(x) and then obviously finding the others will be pretty easy. Thanks in advance!

So far I am stuck because I am attempting to subtract f(x) from f(x+h). No matter how I do it the problem turns out incorrect. I am stuck at -3 square root of x+h - (-3) square root of x.

 

[galactus]

I'll step through it. Follow along. OKey-doke.

The algebra is what tripping you up?.

\(\displaystyle \L\\\lim_{h\to\0}\frac{-3\sqrt{x+h}-(-3\sqrt{x})}{h}\)

\(\displaystyle \L\\\lim_{h\to\0}\frac{-3\sqrt{x+h}+3\sqrt{x}}{h}\)

Multiply top and bottom by the conjugate of the numerator:

\(\displaystyle \L\\\lim_{h\to\0}\frac{(-3\sqrt{x+h}+3\sqrt{x})}{h}\cdot\frac{(-3\sqrt{x+h}-3\sqrt{x})}{(-3\sqrt{x+h}-3\sqrt{x})}\)

The num. reduces down quite nicely:

\(\displaystyle \L\\\lim_{h\to\0}\frac{9\sout{h}}{\sout{h}(-3\sqrt{x+h}-3\sqrt{x})}\)

\(\displaystyle \L\\\lim_{h\to\0}\frac{9}{-3\sqrt{x+h}-3\sqrt{x}}\)

Now, let h approach 0. What do you have?. Bet it's the derivative of \(\displaystyle {-}3\sqrt{x}\).

 

[Mathamateur]

Wow, that made it perfectly clear to me.

I ended up getting -3 over 2 square root of x

Not sure why this one tripped me up so bad, possibly because of the negative numbers in front of the square root, who knows. But anyways, thanks a ton.