Jack Mutchnik 4th Post "Product and Quotient Rule"

Product and Quotient Rule!

Hello class I am Professor Mutchnik and today we will be learning about applying the product and quotient rule to find the derivative of functions.

Lets start with the Product Rule: In calculus the product rule is used to find the derivative of the product of two or more functions 

The product rule states that if U= f(x) and V =g(x) and these two functions are different then;

 f(x) x g(x) = f'(x) * g(x) + f(x) * g'(x)

A simple trick to remembering this is:

(Derivative of the first function) * (the second function) + (the first function) * (the derivative of the second function)

Luckily the product rule is simple and if you accidentally switch the order of the two functions, you will still get the same answer.

Here is a example:

Find the derivative of: y = (x+5)(3x-2)

y' = (1+0) * (3x-2) + (x+5) *(3-0)

y' = (3x-2) + (3x+15)

y' = (6x + 13)

Now you give it a try find the derivative for the function:

y = (6x^2 + 5x +45)(7x + 13)

Know to the Quotient Rule: In calculus the quotient rule is used to find the derivative of the division of one or more functions

If y= f(x)/g(x)

Then y' = f'(x)*g(x) + f(x)*g'(x)

              g(x)^2

A simple trick to remember this is:

y' = Derivative of the top function*bottom function + top function*derivative of the bottom function

Bottom function^2

Heres an Example:

y = (x^2 +1)/(x-1)

y' = (2x)(x-1) + (x^2+1)(1)

(x-1)^2

=(2x^2-2x) + (x^2+1)

(x^2-2x-1)

=  (3x^2-2x+1)

     (x^2-2x-1)

Now you try:

Find the derivative of:

 y=(3x^3+4)/(2x^2+3x+1)

Know that you know the product and quotient rule any problem involving multiplying and dividing functions to find a derivative should be a peace of cake!

Professor Mutchnik Out