Calculate f '(t) directly from the definition of derivative...

[flaren5]

I didn't include all of my work, since I wasn't having difficulty while solving the problem. Where I seem to be having some issues is with the final answer. I got the final answer from the book, but I'm not sure how the final answer was simplified. I'm assuming that the h in the denominator and numerator cancel out? ...but I'm not sure what happen to the 2 in the numerator, or the +1 in the denominator. Any insight to this matter would be greatly appreciated.

 

[Deleted member 4993]

View attachment 3041

I didn't include all of my work, since I wasn't having difficulty while solving the problem. Where I seem to be having some issues is with the final answer. I got the final answer from the book, but I'm not sure how the final answer was simplified. I'm assuming that the h in the denominator and numerator cancel out? ...but I'm not sure what happen to the 2 in the numerator, or the +1 in the denominator. Any insight to this matter would be greatly appreciated.

After cancelling 'h' s out - you take the limit h → 0.

Then the denominator will become 2 *√(2t+1) - and now '2's cancel out.

Good work...

 

[lookagain]

View attachment 3041

laren5, you must have the "limit as h approaches 0" notation in front of your expression in the second line. As it is, that step of yours is incorrect.

\(\displaystyle f'(t) \ = \ \displaystyle\lim_{h \to 0} \ \dfrac{2h}{h(\sqrt{2(t + h) + 1 \ } + \sqrt{2t + 1 \ } \ )} \)