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The Quotient Rule - Problem 2
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The quotient rule states that the derivative of a function h(x) where h(x) = f(x)/g(x) is h'(x) = (g(x)f'(x) - f(x)g'(x))/(g(x))^{2}.

For example, let h(x)=x^{2}/4x^{3}-7. Our functions f and g are f(x)=x^{2} and g(x)=4x^{3}-7. By using normal differentiating rules, we know that f'(x)=2x and g'(x)=12x^{2}. Plug these values into the formula: h'(x)=((4x^{3}-7)(2x) - (x^{2})(12x^{2}))/(12x^{2})^{2}. By expanding, we get h'(x)=(8x^{4}-14x - 12x^{4})/144x^{4} = (-4x^{4}-14x)/144x^{4}. Simplified, this becomes h'(x)=(-2x^{3}-7)/72x^{3}.

Let's do another quotient rule problem. Here we're asked to differentiate y equals e to the x minus 1 over e to the x plus 1. So the numerator and the denominator are very similar, but this is still an interesting function to look at if you graph it.

The derivative, dy/dx, is going to be, and I have to use the quotient rule for this guy. Here is my low function, here is my high function. It's going to be low d high e to the x times the derivative of e to the x minus 1, which is e to the x, minus high d low. So e to the x minus 1 times the derivative of this guy which is also e to the x over, the square of what's below. So e to the x, now I'm remembering plus in the denominator, plus 1.

I can simplify this a little bit. This becomes, I have to the x times e to the x, which is e to the x quantity squared, or e to the 2x plus e to the x minus. Then I have e to the x times e to the x which is another e to the 2x, so I can see that those are going to cancel. Then minus, minus gives me +1 times e to the x. So plus to the x. And the denominator e to the x plus 1². So these guys cancel. These two add up to 2e to the x. My final answer is 2e to the x over e to the x plus 1 quantity squared. That's my derivative.

Let's do another one. I have y equals 6 root 6 over x plus 2. I want dy/dx. So I'm going to use the quotient rule on this. Here is my low function, here is my high function. Remember the derivative of the square root of x is 1 over 2 root x. Let me just write this on the side, the derivative. You get that by thinking of the square root of x as x to the ½. So that's ½x to the -½. X to the -½ is the same as 1 over the square root of x. So this is ½ times 1 over root x. That's the same as 1 over 2 root x. So a little side bar here. We'll come back to that in a second.

Low d high, x plus 2 times the derivative of this function. So the derivative of 6 root x is going to be 6 times this. So 3 over root x minus high d low. So 6 root x times the derivative of x plus 2 which is 1, over the square of what's below. So this is a little complicated, because we've got a tiny fraction here.

Let's see what we can do about that. Actually one thing that you can do about that to get rid of fractions within fractions, these are called complex fractions, is you can multiply the top and bottom of this answer with root x. I'm going to do that. Your teacher may want you to do that. Simplifying complex fractions is something that many students are required to do.

So multiplying the root x through the numerator, you get a cancellation with this guy. You'll have 3 times x plus 2, so that's 3x plus 6 minus. Here root x times root x is just x. So you'll have minus 6x. You do get a root x in the denominator, but that's okay. I think this looks nicer than the one we had before. Simplifying the numerator, you get -3x plus 6 over root x times x plus 2 quantity squared.

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