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Find the derivative of the function using the definition of a derivative. State the domain of the function and the domain of its derivative.$g(x)=\sqrt{1+2 x}$

domain of the function $\left[-\frac{1}{2}, \infty\right)$ddomain of the function $\left(-\frac{1}{2}, \infty\right)$

Calculus 1 / AB

Chapter 3

Derivatives

Section 2

The Derivative as a Function

Villele V.

September 30, 2020

how did we get a 2 at the denominatr?

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Okay, so here we want to use the definition of the derivative which states that the definition of a derivative where the function is f of X, that's gonna be equal to the limit as H approaches zero of the function evaluated at express age minus the function evaluated at X over H. And so for prom 27 um, defines our function is G of X vel McGee is sort of relevant being equal to the square root of one plus two x. And so this means that if we evaluate this function at X plus H, that's going to be equal to the square root of one plus two times X plus h. We can plug both of these into our formula here, and we will get the limits as H approaches zero of the square root of one plus two x plus two age. So here just distributed that too minus the square root of one plus two x over h okay. And so that we can multiply both the top and the bottom. Bye. The square root of one plus two x plus two h plus the square root of one plus two X is gonna be to try get rid of some of these square roots. Can we have to do the same in the bottoms that we're really just multiplying it by one. And once we do that, we end up with the limit as age of purchase zero of one plus two x plus two H minus one minus two X over h times the square root of one plus two x plus two h plus thes queer of one plus two acts. Okay. And then we get some cancellations. Right. So, uh, the one and the two X will both go away. Yes, we have just the two h on top. And so the H is going to cancel with the age below. And so this is equal to the limit. As H approaches zero of two over in course of H has had to zero this to H is also gonna go away. So we get to over two times the square root. So times the square one plus two x and I have skipped a little step here. Right? So we're evaluating it at H equals zero house. We could remove this limit piece. Okay. And then the twos, they're gonna cancel out, right? So we get that these twos cancel. And so the answer. Then it's just gonna be one over the square root of one plus two x. Okay. And so now this one, right? We can't divide by zero. So we do, in fact, have not a full domain. And in fact, we can't. We can't take, um, the square root of a negative number if we're gonna work in reald numbers and so ex would need to be greater than negative 1/2 for the derivatives of the domain of the derivative is going to be negative 1/2 or larger, and then for the the function g of X. Well, this time we're not worried about dividing by zero, but we still don't want to be negative in the square root. So the domain here is gonna be negative 1/2 but including the negative 1/2 because it can equal zero ah to infinity.

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