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(s) = 4s³+3s; a= -3, -1

Accepted Answer

Hello. In this video, we are going to be determining the derivative of the function using the limits. So the function given to us is G of T. Equal to 5 T squared plus 2 T. We want to go ahead and find the derivative by using limits. In order to use limits to find the derivative, the one tool that we have in hand is the definition of a derivative. The definition of the derivative is as follows. We have the limit as H approaches 0 of G of T plus H minus G of T is all divided by H. So what we want to go ahead and do is you want to use this definition of the derivative in order to find the derivative using limits. Now, this definition requires two things. First, we need the GFT function. That is already given to us as it is the original function that we are trying to find the derivative to. But next, what we need is the GFT + H function. We can go ahead and solve this by plugging in the quantity T + H into the original function. That will give us G of T + H equal to 5 multiplied by T + H quantity squared, plus 2 multiplied by T + H. Let's go ahead and simplify this function. First, we will go ahead and expand T + H quantity squared. That will give us 5 multiplied by the quantity, T2 plus 2 TH plus H squared. Then we will go ahead and distribute 2 into the quantity T plus H. That will give us 2 T plus 2 H. What we can go ahead and do next is distribute 5 into the quantity T2 + 2 TH plus H2. That will give us 5 T 2 plus 10 TH plus 5 H2 + 2 T + 2 H. And this will be the simplest form of our G of T plus H function. Now that we have the two pieces of information we need, let's go ahead and plug this in to the definition of a derivative in order to find the derivative using limits. That is going to give us the limit, as H approaches 0 of G T plus H, which is 5 T 2 plus 10 TH plus 5 H quad, plus 2 T. Plus 2 H minus the original G of T function which was given to us as 5 T squared. Minus Sorry, plus 2 T. and this will all be divided by the variable H. Now what we want to go ahead and do is simplify the numerator. The first thing we can do is we can distribute -1 into the quantity 5T 2 minus 2T. This will allow us to change the signs of these terms, leaving us with negative 5T 2 minus 2T. If we combine like terms, we can zero out 5 T 2d and negative 5T 2. We may also zero out -2 T with positive 2T. This will simplify the limit to be the limit as H approaches 0 of 10 TH. Plus 5 H squared, plus 2 H, all divided by H. Now, our goal here is to eliminate the H in the denominator. Notice that every term in the numerator has at least one multiple of H, which this means is that we can go ahead and factor out an H from the numerator. This will allow us to rewrite the limit to be the limit as H approaches 0 of H multiplied by the quantity, 10 T plus 5 H plus 2 all divided by H. We can go ahead and eliminate the H's from the numerator and denominator, and this will further simplify the limit to be the limit. As H approaches 0 of 10 T plus 5 H + 2. This is the simplest form of the limit, so now what we can go ahead and do is evaluate the limit by using direct substitution. We will apply the value of the limit H approaches 0 to the variable H. This will give us 10 T + 5 multiplied by 0 + 2. which will further simplify to be 10 T + 2, and this is going to be the derivative G of T. By the use of the definition of the derivative. With that being said, the solution to this problem is going to be B. So I hope this video helps you in understanding on how to use the definition of a derivative, and I will go ahead and see you all in the next video.

21–30. Derivativesa. Use limits to find the derivative function f' for the following functions f.f(s) = 4s³+3s; a= -3, -1

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21–30. Derivatives
a. Use limits to find the derivative function f' for the following functions f.
f(s) = 4s³+3s; a= -3, -1