The Derivative as a unction

Raimuna · #1 $Old$ October 25th, 2009, 07:23 PM

If f(t)= √((t^2-1)/(t^2-1))
Find f '(t)
I have tried it several times using f(x+h)-f(x) / h, but I always get lost and don't know how to solve it.
Could someone help me?
Thank you

ramiee2010 · #2 $Old$ October 25th, 2009, 07:41 PM

Quote:

Originally Posted by Raimuna $View Post$

If f(t)= √((t^2-1)/(t^2-1))
Find f '(t)
I have tried it several times using f(x+h)-f(x) / h, but I always get lost and don't know how to solve it.
Could someone help me?
Thank you

are you sure it is $f(t)=\sqrt {\frac{t^2-1}{t^2-1} }$ ?

adkinsjr · #3 $Old$ October 25th, 2009, 07:44 PM

Quote:

Originally Posted by Raimuna $View Post$

If f(t)= √((t^2-1)/(t^2-1))
Find f '(t)
I have tried it several times using f(x+h)-f(x) / h, but I always get lost and don't know how to solve it.
Could someone help me?
Thank you

That's the function?

$f(t)=\sqrt{\frac{t^2-1}{t^2-1}}=1$

??

It's probably this:

$f(t)=\frac{\sqrt{t^2-1}}{t^2-1}$

right?

Raimuna · #4 $Old$ October 25th, 2009, 07:44 PM

Quote:

Originally Posted by ramiee2010 $View Post$

are you sure it is $f(t)=\sqrt {\frac{t^2-1}{t^2-1} }$ ?

sory, my bad. the denominator should be t^2+1

adkinsjr · #5 $Old$ October 25th, 2009, 07:56 PM

So we want to evaluate the limit:

$\lim_{h->0}(\frac{\sqrt{(t+h)^2-1}}{h(t+h)^2-h}-\frac{\sqrt{t^2-1}}{ht^2-h})$

aukie · #6 $Old$ October 25th, 2009, 08:29 PM

Why dont you just use the rules of differntiation?
I dont see why you would try to find the derivative of f using the limit definition of the derivative.

Two obvious approaches that stand out.

first you could use the chain rule followed by the quotient rule for differentiation.

or simplify

$f(t)=\sqrt {\frac{t^2-1}{t^2 + 1} } = \frac{\sqrt {t^2-1}} {\sqrt{t^2 + 1}}$

then use the quotient rule and chain rule.

I find after some quick and sloppy calculations (so forgive me if this is not completely accurate)

$f'(t) = \frac{2t}{(t^2 - 1)(t^2 + 1)^{3/2}}$