21–30. Derivatives
a. Use limits to find the derivative ...

Question

21–30. Derivatives

a. Use limits to find the derivative function f' for the following functions f.

f(t) = 1/√t; a=9, 1/4

Accepted Answer

Welcome back, everyone. In this problem, we want to determine the derivative of the function G of X equals 1 divided by the square root of X + 1 using limits. A says it's -1 divided by 2 multiplied by x + 1 multiplied by the square root of X + 1 D 1 divided by 2 multiplied by X + 1 multiplied by the square root of X + 1. C -1 divided by 2 X multiplied by the square root of X + 1 and D 1 divided by 2 X multiplied by the square root of X + 1. Now, how can we determine the derivative of GFX using limits? Well, we can use the limit definition of the derivative. And basically, what the definition tells us is that for GF X, then the derivative of GFX. Is going to be equal to the limit. As H approaches 0 of G of X plus H minus G of X, OK, all divided by H. So essentially what it means is that if we can find G of X plus H, plug in G of X, then we should be able to use this expression to figure out the derivative of G of X. So let's go ahead and do that, OK? First, let's determine GX plus H. Now to find G X plus H, we're simply putting X + H into G of X wherever we see X. So G X + H will be 1 divided by the square root of X plus H + 1. Now that we have G X plus H, we can substitute it and the G of X into the limit definition. Because no, that means then the derivative of G of X is going to be equal to the limit as H approaches 0 of 1 divided by the square root of X plus H + 1. Minus 1 divided by the square root of X + 1 all divided by H. Now, let's try to simplify the expression by finding a common denominator here, and here when we do that, we're going to have the limit as H approaches 0 of the square root of X + 1 minus the square root of X + H+ 1. All divided by H multiplied by the square of X + 1. Multiplied by the square root of X plus H + 1. Now in this case, to make our problems easier to make it simpler, we can rationalize the numerator by multiplying the numerator and the denominator by the conjugate of X + 1, the conjugate of the square root of X + 1 minus the square root of X + H+1, OK. So now, And that means we could rewrite this as the limit as H approaches 0 of our expression. Let me just copy and paste it here in the interest of time, yeah. So the limit as H approaches 0 of all of this multiplied by the square root of X + 1 plus the square root of X + H+ 1. All divided by the square root of X + 1 plus the square root of X + H+ 1. Now when we simplify this, what is that going to be? Well, we can simplify our numerator using the difference of 2 squares, OK? And when we do that, a numerator is going to become X + 1, OK, minus X + H+ 1. And in our denominator, this is going to become H multiplied by the square root of X + 1 multiplied by the square root of X + H+1, multiplied by our conjugate the square root of X + 1 plus the square root of X + H+ 1. OK. Now, if we, you know, look at our numerator here, we could simplify it even some more because if we distribute our negative, that's gonna be minus X minus H, OK, -1. Let me get rid of our brackets here, OK? So minus X, minus H, -1 for our second term. And know when we do that X minus X is 01 minus 1 is also 0, and no, this is gonna be, are you gonna leave us with the limit as H approaches 0 of negative H, OK, divided by all of these 4 denominator. Again, let me just copy and paste this here in the interest of time, but it's gonna be all of that 4 or denominator, OK? I know what's important is that we notice here age cancers age. When each cancers age. OK, we'll be left with, let me write that down here, the limit. As it approaches 0, OK, of -1. They all divided by square root of X + 1 multiplied by the square root of X plus H +1, multiplied by the square root of X + 1 plus the square root of X + H+ 1. And know that we have that, let's substitute H equals 0 directly into our expression. So no, when we do that, that's gonna be -1. All divided by the square root of X + 1 multiplied by the square root of X + 0 + 1, which is going to be left to the square of X + 1, multiplied by the square rate of X + 1 plus the square rate of X + 0 + 1 which would still leave us with the square root of X + 1, OK. And now we can simplify this even further to be -1, all divided by, no, we have the skirt of X + 1 multiplied by the skirt of X + 1 which is going to be X + 1. And if we're adding the skirt of X + 1 twice, that's gonna be multiplied by 2 multiplied by the skirt of X + 1. And no, we can write this even further to say that the derivative of G of X equals the negative value of 1. Divided by 2 multiplied by x + 1 multiplied by the square root of X + 1. This is the derivative of G of X. If we look back at our answer choices, that tells us then that A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

21–30. Derivativesa. Use limits to find the derivative function f' for the following functions f.f(t) = 1/√t; a=9, 1/4

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21–30. Derivatives
a. Use limits to find the derivative function f' for the following functions f.
f(t) = 1/√t; a=9, 1/4