Derivative problem without rules

[QuarkCharmer]

Homework Statement

Find the derivative of
$$g(t)=\frac{1}{\sqrt{t}}$$
I cannot use any derivative rules on this problem. Just the difference quotient.
It's from the Stewart book (6E, Chapter 3, problem 26).

Homework Equations

Difference Quotient

The Attempt at a Solution

Here is where I am at so far, I know the solution if that helps, which is how I know I am making a mistake somewhere, or I don't know how to proceed.

$$g'(t)=\frac{-1}{2x^(\frac{3}{2})}$$
That is 2x^(3/2), it looks funny in the latex.

Any pointers on what to do differently or what to do next?

 

[LCKurtz]

It would be easier to edit and comment if you typed it in. Anyway, starting with

$$\frac{\frac 1 {\sqrt{t+h}}- \frac 1 {\sqrt{t}}}{h}$$

what you should do immediately is multiply both the numerator and denominator by

$$\sqrt{t+h}\sqrt{t}$$

That will avoid your compound fractions. Then you are ready to multiply by the conjugate surd like you did before but it will be simpler, and it will be ready to let h → 0.

 

[jbunniii]

Your general approach looks fine but you made some mistakes going from line 3 to line 4.

After canceling the h in the numerator and denominator, the denominator should be:

$$\sqrt{t}\sqrt{t+h}\sqrt{t} + \sqrt{t}\sqrt{t+h}\sqrt{t+h}$$

Now let h go to zero and you get the desired answer.

P.S. Be careful about what values h is allowed to take, particularly when t = 0.

 

[QuarkCharmer]

Genius!

I did exactly that, and it whittled down to:

$$\frac{-1}{\sqrt{t}\sqrt{t}\sqrt{t}+\sqrt{t}\sqrt{t}\sqrt{t}}$$

$$\frac{-1}{2\sqrt{t^3}}$$

Which is equal to the solution!

Thank you so much, I would have never figured this one out. I didn't think to multiply the conjugate into the bottom (where the h is).