The Quotient Rule - Problem 1


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)=x2/4x3-7. Our functions f and g are f(x)=x2 and g(x)=4x3-7. By using normal differentiating rules, we know that f'(x)=2x and g'(x)=12x2. Plug these values into the formula: h'(x)=((4x3-7)(2x) - (x2)(12x2))/(12x2)2. By expanding, we get h'(x)=(8x4-14x - 12x4)/144x4 = (-4x4-14x)/144x4. Simplified, this becomes h'(x)=(-2x3-7)/72x3.


Let's do some problems with the quotient rule. So let's just recall that the quotient rule is how we differentiate a quotient of two functions; f and g. We'll call f the high function, and g the low function. Remember that the quotient rule is low d high minus high d low over the square of what's below.

So let's take a look at this problem. We want to differentiate y equals e to the x over x². So dy/dx, the derivative, is going to be low d high. So I need to put my fraction bar in. Low d high would be x² times the derivative of e to the x which is e to the x, minus high d low that's e to the x times the derivative of this guy which is 2x, over the square of what's below. That becomes x to the fourth.

This can be simplified a little bit. I can pull an e to the x out. I'm left with x² minus 2x. Now one of those x's will cancel with the denominator. So this x, this x, and that guy. So I'm left with an x minus 2 all over x³. So there is your answer.

Let's try another one. I have y equals x² plus 3x plus 2 over 5 minus x. Again, the derivative is low d high. So this is the low function. D high would be 2x plus 3 minus high d low. The high function is x² plus 3x plus 2. D low would be -1, the derivative of 5 minus x, -1 over the square of what's below. So 5 minus x².

I can simplify this. I don't want to, it looks terrible. 5 minus x, let's see we're going to get -2x². We're going to get 10x minus 3x. So plus 7x and then we're going to get +50. Then here, we'll get minus, minus plus x². We're going to get plus 3x. We're going to get plus 2 all over 5 minus x². I'll leave the denominator that way.

Just to combine like terms in the numerator; I'm going to have minus x². I'm going to have 10x, and I'm going to have plus 17 all that over (5 minus x)². That's the quotient rule.

differentiation quotient rule product rule derivatives