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How To Take The Derivative Of A Function Using The Power Rule

By Edited Nov 13, 2013 0 0

Derivatives are used quite often in the fields of the Physical and Mathematical sciences to represent many types’ real life applications.  A derivative by its most basic definition is the output of a function evaluated with respect to some defined variable.  One of the most identifiable sets of derivatives we see everyday can be found in the study of Physics.  While not always thought out in terms of functions and variables the concepts of distance, velocity and acceleration are all related.  The variable of distance is the derivative of the function of velocity and in tern the function of velocity is the derivative of acceleration.  And this concept makes sense when we realize that the rate of change that is velocity is evaluated with respect to the distance traveled.  Also the acceleration function is evaluated with respect to the velocity function.


In the study of Calculus, there are various rules of differentiation available that we can use to evaluate these derivatives including the Power, Product, Quotient and Chain rules.  Based upon the algebraic makeup of the function for which the derivative is being found determines when each of these rules can be applied.  The most basic of these rules which can be used to differentiate polynomials is the Power Rule.


The Power Rule which is proven by the Binomial theorem states:


Let F(x) = xA; where A is any positive integer.


F’(x) = d/dx [xA] = AxA-1


If this expression is stated in words, the Power rule says that the derivative of function F (which is variable x raised to the power of A) is A multiplied by the variable raised to the power of A minus 1.


A few examples will clarify this concept.  Let’s solve the following expressions using Power Rule differentiation.


1.)    F(x) = x2 – 5x + 10


      F’(x) = 2x – 5 (Recall that the derivative of a constant value is zero)


2.)    H(x) = (1/x3)[2x3 + 4x2 – 6x + 14] 


Let’s begin by rearranging this polynomial algebraically.


      H(x) = (x-3)[2x3 + 4x2 – 6x + 14]


      H(x) = 2x0 + 4x-1 – 6x-2 + 14 x-3


      H(x) = 2 + 4x-1 – 6x-2 + 14 x-3


Now we will find the derivative.


      H’(x) = -4x-2 + 12x-3 – 42 x-4


3.)    h(x) = 3x2 – 3/x2 + ex


        h’(x) = 6x – (x-2)(3) + ex


            h’(x) = 6x + 6x-3 + ex


As we recognize that the derivative of the natural exponent, ex is ex, we see that this is the correct derivative.




Stewart, James. (2001).  Calculus Concepts And Contexts (5th Ed.).  Pacific Grove, CA: Wadsworth Group.



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