# Question Video: Finding the Derivative of a Power Function Using the Definition of the Derivative Mathematics • 12th Grade

Find the derivative of the function π(π₯) = 2π₯β»Β² using the definition of the derivative.

04:27

### Video Transcript

Find the derivative of the function π of π₯ equals two π₯ to the power of negative two using the definition of the derivative.

In this question, we are given a function π of π₯, and we are asked to find its derivative by using the definition of a derivative. To answer this question, we begin by recalling that we say a function π is differentiable at π₯ sub zero if the limit as β approaches zero of π evaluated at π₯ sub zero plus β minus π of π₯ sub zero over β exists and π₯ sub zero is in the domain of π. We can replace π₯ sub zero with the variable π₯ to find the function defined in terms of a limit, which gives us the derivative of π. This is called the derivative of π and is denoted π prime of π₯.

Applying this definition to the given function π of π₯ gives us that π prime of π₯ is equal to the limit as β approaches zero of two times π₯ plus β to the power of negative two minus two π₯ to the power of negative two all over β. If we try to evaluate this limit by substitution, both the numerator and denominator evaluate to give zero, so we end up with the indeterminate form zero over zero. This means that we need to evaluate this limit using a different method. We can evaluate this limit by rearranging. We will start by taking out the factor of β in the denominator, though we need to be careful since this must stay inside the limit.

We will then rewrite each term as a fraction using the laws of exponents. This gives us the limit as β approaches zero of one over β times two divided by π₯ plus β squared minus two over π₯ squared. We can now combine the two rational functions by using cross multiplication. We obtain the following expression. We can now combine the numerators. However, we want to expand the parentheses in the numerator of the second term first. This gives us two π₯ squared plus four π₯β plus two β squared. Combining these terms then gives us the limit as β approaches zero of one over β times two π₯ squared minus two π₯ squared minus four π₯β minus two β squared over π₯ squared times π₯ plus β squared.

We can now simplify in order to help us evaluate this limit. First, two π₯ squared minus two π₯ squared equals zero. Next, we can see that both terms in the numerator have factors of β. We can cancel factors of β in the numerator and denominator inside a limit since we are not interested in what happens when β equals zero, only when β approaches zero. This gives us the limit as β approaches zero of negative four π₯ minus two β over π₯ squared times π₯ plus β squared. This is a rational function, so we can attempt to evaluate this limit by direct substitution. We want to substitute β equals zero into the function to evaluate this limit. This gives us negative four π₯ minus two times zero all over π₯ squared times π₯ plus zero squared.

We can simplify this expression by noting that two times zero equals zero and π₯ plus zero is just π₯. So we have negative four π₯ over π₯ squared times π₯ squared. At this point, we want to cancel the shared factor of π₯ in the numerator and denominator. And we can do this since π₯ equals zero is not in the domain of π of π₯. So, we already know that π is not differentiable at this value of π₯ since it is not defined here. This then gives us negative four over π₯ cubed. We can use the laws of exponents to rewrite π prime of π₯ in the same form as π of π₯.

Hence, we have shown that if π of π₯ is two π₯ to the power of negative two, then π prime of π₯ is equal to negative four times π₯ to the power of negative three.