Product and Quotient Rule – When L’Hopital’s Fails
Now, in our video over derivative properties and formulas we talk about how when you have a function with terms being added or subtracted, you can take the individual derivative of each terms and just replace them and put the correct signs.
Well, what about when we are asked to take the derivative of a functions with terms being multiplied or divided?
Could we just multiply the terms together then take the derivative of the product? Or could we take the derivative of the two terms being multiplied and then multiply those together?
Well, the answer is no, and I’ll show you why.
Let’s say we want to take the derivative of the product of the following functions: f(x)= x4 and g(x)=x8.
So, let’s try the first method I mentioned. Multiplying these two functions together, then taking the derivative of their product.
(f(x)*g(x))’ = (x^4*x^8)’ = (x^12)’= 12x^11
Now, let’s try the other method I mentioned. Let’s take the derivative of the two terms being multiplied and then multiply them together. For either of these methods to work, we would need to get the same answer using either method.
We can see that these two answers are different. So, we can see that (f*g) does not equal f’ * g’
If we try this using the exact same functions to apply this same concept with quotients, we can see that we will have a similar result.
(f/g)’=(x^4/x^8)’ = (1/x^4)’ = (x^-4)’ = -4x^-5= – 4/x^5
Now if we try taking the derivative of the top and the bottom, then dividing let’s see what we get.
f’(x)/g’(x)= 4x^3/8x^7 = 1/2x^4
Again, we can see that (f/g)’ does not equal f’/g’.
So, in order to take the derivative of functions with products and quotients we will need to use the Product rule and the Quotient rule.
Here is what the product rule states:
If two functions f(x) and g(x) are differentiable (which just means that the derivative exists) then the product is differentiable and,
(f*g)’=f’*g + f*g’
Now, let’s take a look at how to apply this.
Differentiate the following.
5throot(x3) * (3x^2+6x)
So, we can see here that we have two functions that are being multiplied together.
I’m going to go ahead and rewrite this to make it a bit simpler.
Y=x^3/5 * (3x^2+6x)
Now, let’s look at what our product rule is telling us to do. So, we need to take the derivative of the first part of our function, then multiply it by our second part (leaving the second part alone). We then, add the product of the first part of the function left alone by the derivative of the second part.
Alright, so let’s go ahead and do what the product rule is telling us to do.
Y’= (x^3/5)’(3x^2+6x) + (x^3/5)(3x^2+6x)’
So, now we have written out the derivative, but we need to perform our operations. We need to take the derivative where it says to take the derivative, then multiply the two functions together.
Y’= (⅗ x-⅖)(3x^2+6x) + (x^3/5)(6x +6)
Now, we have taken the derivative, but we still need to multiply our functions together here.
Y’= 9/5 x^8/5 + 18/5 x^3/5 + 6x^8/5 + 6x^3/5
We can simplify this even further by adding together our like terms.
Y’= 39/5 x^8/5 + 48/5 x^3/5
And there is our answer!
Alright, now, let’s look at the Quotient Rule.
The Quotient Rule states:
If two functions f(x) and g(x) are differentiable then the quotient is differentiable and,
(f/g)’= (f’*g - f*g’)/(g^2)
Let’s look at how this applies.
Differentiate the following.
So, our Quotient rule tells us to take the derivative of the numerator, multiply it times the denominator left alone, then subtract the product of the numerator left alone by the derivative of the denominator. Then, finally, to divide all of this by the denominator squared.
Let’s start by writing out exactly what the Quotient Rule is telling us to do.
Z’(t)=((2t+8)’(3-9t) – (2t+8)(3-9t)’)/ (3-9t)^2)
Now, let’s perform the operations.
Z’(t) = (2(3-9t) - (2t+8)(-9))/(3-9t)^2
Z’(t) = (6-18t+18t+72)/(3-9t)^2
Z’(t) = (78)/(3-9z)^2
And there is our final answer.
Differentiating products and quotients will become much simpler if you memorize the Product and Quotient Rule. Once you have that down, it’s just a matter of being able to take individual derivatives, and doing some algebra.
I hope this video was helpful. For further help be sure to check out more of our videos by subscribing to our channel below.
See you next time.