21–30. Derivatives
b. Evaluate f'(a) for the given ...

Question

21–30. Derivatives

b. Evaluate f'(a) for the given values of a.

f(x) = 1/x+1; a = -1/2;5

Accepted Answer

Below there today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Given the function H of X is equal to 1 divided by X + 3, find H C for C equals -2 and C equals 2. Awesome. So it appears for this particular problem we're asked to solve for three separate answers. So our first answer, it doesn't directly specify, but in order for us to find H of C for when C equals -2 and 2, we need to first figure out the derivative of H X. So that's our first answer. So our first answer, we need to figure out the derivative of HF X. And once we figure out the derivative of H of X, we can therefore find our last two answers, which is for H C, where C equals -2, which is our second answer, and then our third answer for when C equals 2. So with that said, first off, we need to figure out the derivative of H of X is equal to 1 divided by X + 3. So we need to take this function for H of X and find the derivative, and when we find the derivative, we're going to be finding the value for H X. So that is what we're trying to solve for is we do not know what the derivative is. So we're trying to figure out what H of X is. So in order to do that, we need to recall and use the quotient rule of differentiation. So we need to note that if H of X is equal to U X divided by V of X, then we need to note that H. of X is going to be equal to U X multiplied by V X minus U. Of X multiplied by the of X. All divided by. V of X, all to the power of 2. Fantastic. So, we need to note that when H of X is equal to 1 divided by X + 3, we need to let U X equal 1 and we also need to let V of X equals X + 3. And we also need to note that U of X will be equal to 0, and we need to note that V of X will be equal to 1. So, our next step now is we need to substitute in our values for U of X, V of X, and UX and V X into our quotient rule formula, and we will find that. H X is equal to 0 multiplied by x + 3 minus 1 multiplied by 1, all divided by X + 3 to the power of 2, which will be equal to -1. So you take -1 divided by. X + 3 to the power of 2. So now we can we can now solve for our second answer. So we've we solved one of our answers, we found the derivative. All right, we did it, so. Once again H X is equal to negative 1 divided by X + 32. And to be more specific, you can also write the negative sign out front, so you could say negative, and let's write this a little nicer, you could write negative 1 divided by X + 32. And this is going to be equal to our derivative, which is H X. So that's our first answer. Now we could solve for our second answer, and our second answer once again is for H. Of C Where C in this case is going to be equal to. 2. So this is C is equal to -2. So we're trying to find H C where C equals -2. So when we do that, we will find that when we plug in our value of -2 N for C, we will find that H of -2 is equal to 1 divided by. -2. Plus 3 to the power of 2, which is equal to -1 divided by 1, which is equal to -1, and that is our second answer, and once again, this is for H-2 is equal to -1. Fantastic. And similarly, we can now solve for our third and final answer, H of C, where C is equal to 2, that will mean that H2 is equal to -1 divided by. Positive 2 + 3 to the power of 2, which is going to be equal to. it's going to be equal to drum roll, -1 divided by 25 when you plug that into your calculator, and that is going to be our third and final answer. So this is for when H2. So H2 is equal to -1 divided by 25 or -125. And that's it. Those are our final three answers. Hooray, we did it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video.

21–30. Derivatives
b. Evaluate f'(a) for the given values of a.
f(x) = 1/x+1; a = -1/2;5

Key Concepts

Derivatives

Quotient Rule

Evaluating Derivatives at Specific Points

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