Blog Post #4 --> Ana Maria Lopez

HELLOO!

My name is Professor Lopez and today we will be covering the how to find the derivative using the product and the quotient rule. Even though the term derivative might seem complicated, these two rules will only make the process easier and make a “short-cut”.

Before we start with any new material, I will first like to explain the purpose for this lesson. Not only are derivatives a universal mathematical process, but also they are seen daily in many aspects of life. Derivatives express the slope or the rate of change of a function at a given point. Therefore, knowing how to solve, even when there is a product or a quotient, is essential for life. A great real life example of derivatives is for finding the marginal cost and marginal benefit of a function. This can help out a company know how many extra products they can produce while still producing a profit!! Know that we know the reason behind these rules…

Lets start off with the famous Product Rule:

Lets assume that u= f (x) and v= g (x) and that these two are differentiable function. This would mean that:

(f g)’=f ’g + f g’

Or….

In other words, it means that the derivative of a product is the derivative of the fit times the second plus the fist times the derivative of the second.

Example:

F (x)= x⁴  * e^3x

à d/dx (x⁴  * e^3x ) = d/dx (x⁴ ) * e^3x   + x⁴  d/dx(e^3x)

                                   = (4x) e^3x  + x⁴ (2 e^3x)

                                   =  4x e^3x + 2x⁴ e^3x

Even though the initial problem had many aspects to it, we could easily solve it by just going one step at a time. We first multiplied the derivative of the X⁴ times the e^3x and then added up the X⁴ times the derivative of the e^3x.

Now, lets continue with the quotient rule:

Letts assume that u= f (x) and v= g (x) and that these functions are differentiable. This would mean that:

In other words, this means that the derivative of a quotient is the derivative of the numerator times the denominator minus the numerator times the derivative of the denominator, all over the denominator squared.

Example:

Now that we know how to tackle larger and more “ complex” problems, we can understand econ, physics and even chemistry better!!!