# Using the definition of derivative...

• Mar 17th 2010, 05:09 PM
KarlosK
Using the definition of derivative...
This is going to be an easy one

but I have to solve the square root of (2x+1) using the definition of limit.

I set the problem up f(x) lim h-> 0 sq root (2x + h + 1) - sq root (2x + 1) / h then I multiplied by the conjugate.

That gave me (2x + h + 1) - (2x + 1) / h (sq root 2x + h + 1) + (sq root 2x+1)

I am not sure where to go from here. How do I simplify it? I know the answer I just don't see how to get there. Thanks!!
• Mar 17th 2010, 05:15 PM
skeeter
Quote:

Originally Posted by KarlosK
This is going to be an easy one

but I have to solve the square root of (2x+1) using the definition of limit.

I set the problem up f(x) lim h-> 0 sq root (2x + h + 1) - sq root (2x + 1) / h then I multiplied by the conjugate.

That gave me (2x + h + 1) - (2x + 1) / h (sq root 2x + h + 1) + (sq root 2x+1)

I am not sure where to go from here. How do I simplify it? I know the answer I just don't see how to get there. Thanks!!

$\displaystyle \lim_{h \to 0} \, \frac{\sqrt{2(x+h)+1} - \sqrt{2x+1}}{h}$

$\displaystyle \lim_{h \to 0} \, \frac{2(x+h)+1 - (2x+1)}{h \sqrt{2(x+h)+1} + \sqrt{2x+1}}$

combine like terms in the numerator ... one h in the numerator and denominator will divide out ... then determine the limit.
• Mar 17th 2010, 05:25 PM
KarlosK
Quote:

Originally Posted by skeeter
$\displaystyle \lim_{h \to 0} \, \frac{\sqrt{2(x+h)+1} - \sqrt{2x+1}}{h}$

$\displaystyle \lim_{h \to 0} \, \frac{2(x+h)+1 - (2x+1)}{h \sqrt{2(x+h)+1} + \sqrt{2x+1}}$

combine like terms in the numerator ... the h's in the numerator and denominator will divide out ... then determine the limit.

I guess what I am having trouble with is how to cancel out the square roots on the bottom of the fraction.
• Mar 17th 2010, 05:37 PM
skeeter
Quote:

Originally Posted by KarlosK
I guess what I am having trouble with is how to cancel out the square roots on the bottom of the fraction.

do not mess with the radicals in the denominator, they are part of the solution.

follow my previous directions as given.
• Mar 17th 2010, 05:44 PM
KarlosK
Quote:

Originally Posted by skeeter
do not mess with the radicals in the denominator, they are part of the solution.

follow my previous directions as given.

2h / (bottom part)

I guess you could cancel the H's and have

h/ (bottom part)

but the answer to the problem is 1/ sq root (2x + 1)
• Mar 17th 2010, 05:47 PM
skeeter
Quote:

Originally Posted by KarlosK