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Chain, Product and Quotient Rule: Exercise 2

by Batool Akmal
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      DLM Quotient Rule, Chain Rule and Product Rule Exercise Calculus Akmal.pdf
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    Let?s have a look at our second 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. Now you are 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 two things are dividing. Also recall the rule for differentiation so dy/dx, according to the quotient rule, is vdudx minus udvdx, over v squared, 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. V is x squared, plus 2x, plus 1. When you differentiate 1 or any number, it?ll just go to zero. Remember what we just said constants just disappear. They just go to zero. So this might look a bit complicated but in fact, it makes the entire solution a lot easier. Let?s continue. We now differentiate v dv/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/dx. When I multiply v with u that gives me zero multiplied by x squared plus 2x plus 1 minus and we now have 1 multiplied...

    About the Lecture

    The lecture Chain, Product and Quotient Rule: Exercise 2 by Batool Akmal is from the course Quotient Rule, Chain Rule and Product Rule.


    Author of lecture Chain, Product and Quotient Rule: Exercise 2

     Batool Akmal

    Batool Akmal


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