# Finding the derivative of

1. May 9, 2005

### m0286

I am trying to do a question for calculus. I am supposed to find a minimum time for something.. I know how to do the question but I am stuck on how to find the derivative of this function it really confuses me.

(sq. root 1^2+x^2)/3 + 3-x/5
I need to find dt/dx... and I am clueless on how to find this???
THANKS TONS!

2. May 9, 2005

### robert Ihnot

$$d\sqrt{Y}=\frac{dY}{2\sqrt{Y}}$$ Now for Y = A+B, the rule is:

Y'=A'+B'. In the case of a form such as S/T, you can employ the rule:

d(S/T) =$$\frac{S'T-ST'}{T^2}$$

However, in the problem above we don't need to use that since: d(X/3) is just X'/3. Again in the case

$$\sqrt{\frac{1+x^2}{3}}=\frac{\sqrt{1+x^2}}{\sqrt3}$$ which is simpler to work with.

Last edited: May 9, 2005
3. May 10, 2005

### James R

To differentiate

$$t= \frac{\sqrt{1 + x^2}}{3}$$

put $u=1 + x^2$, then use the chain rule:

$$\frac{dt}{dx} = \frac{dt}{du} \frac{du}{dx}$$

Last edited: May 10, 2005
4. May 10, 2005

### Jameson

Or pulling out constants might make it seem simpler.

$$\frac{1}{3}*\frac{d\sqrt{1+x^2}}{dx}$$ etc...

5. May 10, 2005

### HallsofIvy

Staff Emeritus
Most people find it easier to think of "square root" as "1/2 power". That is,
To find the derivative of $$\frac{\sqrt{1+x^2}}{3} + 3- \frac{x}{5}[/itex], write it as [tex]\frac{1}{3}(1+ x^2)^{\frac{1}{2}}+ 3- \frac{x}{5}[/itex] Then the derivative is [itex]\frac{1}{3}\frac{1}{2}(1+ x^2)^{-\frac{1}{2}}{2x}- \frac{1}{5}$$

Last edited: May 10, 2005
6. May 10, 2005

### m0286

You Guys Are Awesome Thanks Sooo Much!!!