- #1
JayDub
- 30
- 0
This is not homework but it is more practise for me. We have to find the derivative of f(x) = sqrt(2x + 5) using the definition of a derivative, I used to remember how to do this but now I am not able to figure out what I am doing wrong.
Here is what I have.
y' = lim(h -> 0) [sqrt(2{x + h} + 5) - sqrt(2x + 5)] / h
y' = (lim(h -> 0) [sqrt(2{x + h} + 5) - sqrt(2x + 5)] / h) * ([sqrt(2{x + h} + 5) + sqrt(2x + 5)] / [sqrt(2{x + h} + 5) + sqrt(2x + 5)])
y' = lim(h -> 0) [2(x + h) + 5 - (2x + 5)] / h[sqrt(2{x + h} + 5) - sqrt(2x + 5)]
y' = lim(h-> 0) [2x + 2h + 5 - 2x - 5] / h[sqrt(2{x + h} + 5) - sqrt(2x + 5)]
y' = lim(h ->0) 2h / h[sqrt(2{x + h} + 5) - sqrt(2x + 5)]
y' = lim(h->0) 2 / [sqrt(2{x + h} + 5) - sqrt(2x + 5)]
I have no idea how to format the questions to make them look nice.
Here is what I have.
y' = lim(h -> 0) [sqrt(2{x + h} + 5) - sqrt(2x + 5)] / h
y' = (lim(h -> 0) [sqrt(2{x + h} + 5) - sqrt(2x + 5)] / h) * ([sqrt(2{x + h} + 5) + sqrt(2x + 5)] / [sqrt(2{x + h} + 5) + sqrt(2x + 5)])
y' = lim(h -> 0) [2(x + h) + 5 - (2x + 5)] / h[sqrt(2{x + h} + 5) - sqrt(2x + 5)]
y' = lim(h-> 0) [2x + 2h + 5 - 2x - 5] / h[sqrt(2{x + h} + 5) - sqrt(2x + 5)]
y' = lim(h ->0) 2h / h[sqrt(2{x + h} + 5) - sqrt(2x + 5)]
y' = lim(h->0) 2 / [sqrt(2{x + h} + 5) - sqrt(2x + 5)]
I have no idea how to format the questions to make them look nice.