# Help with derivative of

1. Sep 30, 2004

### buffgilville

Derivative of square root:
1) Use derivative rule (f(x) = f(x+h) - f(x) / h) to find square root of (6x-4)

I got: 6 / ((square root of 6x+6h-4) + (square root of 6x-4))
but is this in the simpliest form? I thought you can't have a square root in the denominator.?.

2) (derivative) 3 / (x+1)

I got 1 but the I got it wrong and I tried redoing it but still got it wrong. Can someone please help?

2. Sep 30, 2004

### ShawnD

Maybe it's just me, but this question doesn't seem to make any sense at all.

First, check for the answer doing the easy way.
You're given this:

$$3(x + 1)^{-1}$$

so the derivative is this:

$$3(-1)(x + 1)^{-2}(1)$$

$$\frac{-3}{x^2 + 2x + 1}$$

Now to actually do the problem the way you're supposed to, it's just a matter of add/subtracting fractions

$$\frac{\frac{3}{x + h + 1} - \frac{3}{x + 1}}{h}$$

$$\frac{3(x + 1) - 3(x + h + 1)}{(x + h + 1)(x + 1)(h)}$$

If you keep working at it, you eventually will get the right answer
(and yes, I actually did get to the answer working from the formula above)

Last edited: Sep 30, 2004
3. Sep 30, 2004

### HallsofIvy

Staff Emeritus
Perhaps you meant find the derivative of square root of (6x-4)?

Your main problem is that in "6/((square root of 6x+6h-4) + (square root of 6x-4))" you've left out an important part: the limit as h-> 0.

if f(x)= √(6x-4), then f(x+h)= √(6x+ 6h-4) and so the "difference quotient" is $\frac{\sqrt{6x+6h-4}-\sqrt{6x-4}}{h}$. Since you want to be able to cancel that h in the denominator try "rationalizing the numerator" by multiplying numerator and denominator by $\sqrt{6x+6h-4}+\sqrt{6x-4}$. In the numerator we get the "difference of two squares", 6x+6h-4- 6x+4= 6h. In the denominator we have the complicated $h\sqrt{6x+6h-4}+\sqrt{6x-4}$ but that's okay: the difference quotient is now $\frac{6h}{h\sqrt{6x+6h-4}+\sqrt{6x-4}}$ and we can cancel the "h"s to get $\frac{6}{\sqrt{6x+6h-4}+\sqrt{6x-4}}$. Now take the limit as h-> 0 to get $\frac{6}{2\sqrt{6x-4}}$.

f(x)= 3/(x+1) so f(x+h)= 3/(x+h+1) and the difference quotient is $\frac{\frac{3}{x+h+1}-\frac{3}{x+1}}{h}$. To subtract those fractions in the numerator, you have to get the common denominator: x+1 and x so we have $\frac{\frac{x+1- (x+h+1}{x(x+h+1}}{h}$
That gives $\frac{-h}{hx(x+h+1}= \frac{-1}{x(x+h+1)}$
Now take the limit as h goes to 0.