Chain, Product and Quotient Rule: Exercise 2

00:00 Let's have a look at our 2nd question. We are now looking at a function that is being has a number and has a function dividing it. Let's just discuss the kind of things we can do here. So we have y equals to 1 over x squared plus 2x plus 1.

00:19 Now, you're often able to bring the whole function up if you ever wanted to and just use the product rule. However in this case, we'll just stick to the quotient rule because it's probably the easiest way to do it. You can say that the top function is u and the bottom function is v. Remember, we apply the quotient rule when we notice that 2 things are dividing. Also recall the rule for differentiation, so dy by dx according to the quotient rule is vdudx minus udvdx over v squared.

00:52 So we need all of these components to put them into this formula. If we look at u now, we have 1 so it's not actually a function of x, but we'll see what we can do with it.

01:02 V is x squared plus 2x plus 1. When you differentiate 1 or any number, it'll just go to 0. Remember what we said, constants just disappear, they just go to 0.

01:17 So this may look a bit complicated but in fact it makes the entire solution a lot easier.

01:22 Let's continue. We now differentiate vdv by dx or v dash, that gives you 2x plus 2 and obviously that one just disappears. We can now apply the quotient rule that says vdudx minus udvdx divided by v squared. So we have dy by dx. When I multiply v with u, that gives me 0 multiplied by x squared plus 2x plus 1 minus and we now have 1 multiplied by 2x plus 2 all over v squared. So, x squared plus 2x plus 1 all squared at the bottom. So the good thing that happens here, anything multiplied with 0 just becomes 0. So that term will just disappear, leaving us with just minus 2x plus 2 because 1 doesn't bother us over x squared plus 2x plus 1 all squared. Now we can simplify the equation even further by factorizing the numerator and denominator as we can factor the denominator into x plus 1 to the power of 4 and the numerator into minus 2 times x plus 1. The x plus 1 in numerator and denominator cancel out. This leaves us with the final answer in its simplest form, minus 2 divided by x plus 1 to the power of 3.