Using the quotient rule, 2016 Calculus AB free response 6b
About this video
Sal solves 2016 AP Calculus AB 6b: finding H'(1) where H(X) = G(X)/F(X), using values from a table. He applies the quotient rule, H' = (G'F - GF')/F^2, with G'(1) = 8, F(1) = -6, G(1) = 2 and F'(1) = 3 to get (-48 - 6)/36 = -54/36 = -3/2.
You'll be able to
- state the quotient rule for H(X) = G(X)/F(X).
- substitute table values into the quotient rule to find a derivative at a point.
- simplify the result to lowest terms.
Key words
- Quotient rule
- The derivative of G/F is (G'F - GF')/F^2: derivative of the top times the bottom, minus the top times derivative of the bottom, over the bottom squared.
- Numerator
- The top function of a fraction, here G(X).
- Denominator
- The bottom function of a fraction, here F(X).
Transcript
- [Instructor] We're told the functions F and G have continuous second derivatives. The table above gives the values of the functions and their derivatives at selected values of X. Let H of X be equal to G of X over F of X. Find H prime of one. Pause this video, have a go at that.
All right, now let's just do this together. So if we're trying to find the derivative of H of X at one, and they don't have anything about H of F here, but they give us F of X and F prime of X at different values of X, this looks pretty clear that we should be using the quotient rule.
The quotient rule tells us that H prime of X, and this is really just me spouting out the quotient rule, which we prove in other places in Khan Academy, it's going to be equal to G prime of X, the derivative of the top one, times the second function, or the function in the denominator I should say, derivative of the function in the numerator times the function in the denominator minus the function in the numerator times the derivative of the function in the denominator, all of that over the function in the denominator, the function in the denominator squared.
So if we wanna find the derivative of H at X equals one, we just replace all of the X's with one here. So we want G prime of one times F of one minus G of one times F prime of one. All of that over F of one squared. All right, now what is G prime of one? The derivative of G at X equals one is eight.
So this is eight. What is F of one? F of one, when X equals one, X is negative... When X equals one, F of X is negative six. So this is negative six. And F of one is negative six, but we square it, so this whole thing is going to be equal to 36. G of one, G of one is two.
This is two. And then F prime of one, that is equal to three. That's equal to three. So we get, I'm gonna go all the way over here. When you're actually doing this on the AP exam, you'd have more space. But eight times negative six is -48. Minus two times three, so minus six.
All of that over 36. This would be -54. - 54 over 36. Now let's see, both the numerator and the denominator is divisible by nine. So this is equal to negative six over four. Actually, they were both divisible by 18. But now we can divide by two again. So this is negative three halves.
And we are done.
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