Video: The Power Rule of Derivatives

In this video, we will learn how to use the power rule of derivatives and the derivative of a sum of functions to find the derivatives of polynomials and general power functions.

17:48

Video Transcript

In this video, we will introduce and learn how to use the power rule of derivatives. This will enable us to differentiate power functions with positive, negative, and fractional exponents. We’ll also discuss the rules of finding the derivative of a sum or difference of terms, which will enable us to find the derivatives of polynomial functions with real exponents. We’ll then apply these rules to a series of examples.

Let’s begin by recalling the form or definition of the derivative of a function. For the function 𝑓 of π‘₯, its derivative, denoted as 𝑓 prime of π‘₯, is given by the limit as β„Ž tends to zero of 𝑓 of π‘₯ plus β„Ž minus 𝑓 of π‘₯ over β„Ž at the points where this limit exists. Where does this definition come from? Well, let’s consider the graph of a function 𝑦 equals 𝑓 of π‘₯ whose derivative we want to find at a particular point with the general π‘₯-coordinate π‘₯. If we take another point a short distance away with π‘₯ coordinate π‘₯ plus β„Ž, then we can sketch in a chord connecting these two points.

The slope of this chord will give an approximation for the slope of the curve, and hence the first derivative of the curve, at the point with π‘₯-coordinate π‘₯. We can sketch in a right triangle below this chord and then find an expression for its slope using change in 𝑦 over change in π‘₯. The change in 𝑦 will be the function evaluated at π‘₯ plus β„Ž minus the function evaluated at π‘₯. That’s 𝑓 of π‘₯ plus β„Ž minus 𝑓 of π‘₯. And the horizontal change will be π‘₯ plus β„Ž minus π‘₯. Of course, the π‘₯s in the denominator cancel. And we’re just left with 𝑓 of π‘₯ plus β„Ž minus 𝑓 of π‘₯ over β„Ž. Remember, though, that this is the slope of this short chord, not the slope of the curve itself. Although if β„Ž is sufficiently small, then it will give a good approximation.

What we do then is we consider a sequence of chords that get progressively shorter. So we reduce the value of β„Ž so that these two points get closer together. As β„Ž tends to zero, this sequence of chords will tend to the tangent to the graph of 𝑓 at point π‘₯. The slope of the tangent then β€” which, remember, is the slope of the curve and its first derivative, 𝑓 prime of π‘₯ β€” is the limit as β„Ž tends to zero of this quantity, 𝑓 of π‘₯ plus β„Ž minus 𝑓 of π‘₯ over β„Ž. Which is what we wrote down earlier in our formal definition.

So that’s great. We know the formal definition for finding the derivative of a function. And we know where it comes from. But, in practice, using this definition can be a time consuming and tedious process. So it would be great if there were some general rules that we could apply instead. First, though, let’s use this definition to find an expression for the derivative of the function 𝑓 of π‘₯ equals π‘₯ squared, which is called differentiating from first principles. And then, we’ll use this to see if we can generalize.

From first principles then, 𝑓 prime of π‘₯, for this function 𝑓 of π‘₯, will be the limit as β„Ž tends to zero of π‘₯ plus β„Ž all squared minus π‘₯ squared over β„Ž. Distributing the parentheses will give the limit as β„Ž tends to zero of π‘₯ squared plus two π‘₯β„Ž plus β„Ž squared minus π‘₯ squared all over β„Ž. Now, the π‘₯ squared terms in the numerator will cancel. And then, all of the remaining terms have a common factor of β„Ž which we can simplify by. We’re left with the limit as β„Ž tends to zero of two π‘₯ plus β„Ž. And as β„Ž tends to zero, this limit will be equal to two π‘₯.

So using differentiation from first principles, we’ve found that the derivative of π‘₯ squared is two π‘₯. We could apply exactly the same method to find the derivatives of π‘₯ cubed, π‘₯ to the fourth power, and so on. And if we did, we’d find that the derivative of π‘₯ cubed is three π‘₯ squared and the derivative of π‘₯ to the fourth power is four π‘₯ cubed. Can you spot a pattern in these results? Well, we may spot it more easily if we rewrite two π‘₯ as two π‘₯ to the power of one. What we should notice is that in the derivative, the exponent has been reduced by one each time. And the function has been multiplied by the original exponent.

So, for example, in the case of π‘₯ cubed, the exponent has reduced by one, from three to two. And we’ve multiplied by the original exponent of three. This gives us our first general rule then, which is called the power rule. And it tells us that the derivative with respect to π‘₯ of π‘₯ to the 𝑛th power is equal to 𝑛 multiplied by π‘₯ to the power of 𝑛 minus one. We reduce the exponent by one and multiply it by the original exponent.

Now, it’s important to note that we have only demonstrated this general rule. We haven’t proven it. To do so, at least for positive integer values of 𝑛, would require an application of the binomial theorem. But it’s beyond the scope of this video to look at that here. This rule actually works for any real value of 𝑛. So that includes positive, negative, and fractional exponents.

So that’s great if the coefficient of our term is just one. But what about if we have a general coefficient π‘Ž? So we have a constant multiplied by π‘₯ to some real exponent 𝑛. Now, we could return to first principles, perhaps using the example function 𝑓 of π‘₯ equals four π‘₯ square. If we did so, and you can pause the video and check through the algebra here if you wish, we’d find that 𝑓 prime of π‘₯ is equal to eight π‘₯. Now, how does this link to what we’ve seen the derivative of the simpler function π‘₯ squared to be? Well, eight π‘₯ is equal to four multiplied by two π‘₯. And two π‘₯ is the derivative with respect to π‘₯ of π‘₯ squared.

So we’ve found that by multiplying by a constant, in this case four, the derivative is also multiplied by that same constant. So we can generalize our power rule a little further and say that the derivative with respect to π‘₯ of π‘Ž multiplied by π‘₯ to the 𝑛th power for real constants π‘Ž and 𝑛 is equal to π‘Ž multiplied by 𝑛π‘₯ to the power of 𝑛 minus one. We can also interpret this in practical terms. If we recall that multiplying by a constant achieves a vertical stretch with a scale factor of that constant. So the graph of 𝑦 equals four π‘₯ squared is a vertical stretch of the graph of 𝑦 equals π‘₯ squared by a scale factor of four.

The slope of the tangents to these curves and therefore their derivatives for any given π‘₯-value will be four times as much on the graph of 𝑦 equals four π‘₯ squared compared to what it was on the graph of 𝑦 equals π‘₯ squared. And so, we have an extra factor of four in our derivative or, more generally, an extra factor of π‘Ž.

One further application of the power rule is the rule for finding the derivative of a constant. Suppose you want to find the derivative of the function 𝑓 of π‘₯ equals seven. Well, recall that we can also think of seven as seven π‘₯ to the power of zero, as π‘₯ to the power of zero is one. Applying the power rule would give that the derivative 𝑓 prime of π‘₯ is equal to seven multiplied by zero π‘₯ to the power of zero minus one, which is just equal to zero. So the derivative of our function 𝑓 of π‘₯ equals seven and, in fact, the derivative of any constant 𝑐 with respect to π‘₯ is just zero.

We can, again, see the practical interpretation of this if we consider the graph of 𝑓 of π‘₯ equals seven. It’s a horizontal line through seven on the vertical axis. If we recall that a derivative gives the rate of change of 𝑦 with respect to π‘₯, then we see that as 𝑦 isn’t changing, its rate of change will be zero. In the same way, if we were to consider the graphs of 𝑦 equals 𝑔 of π‘₯ and 𝑦 equals 𝑔 of π‘₯ plus 𝑐 for some function 𝑔 of π‘₯ and some constant 𝑐. We know that the graph of 𝑦 equals 𝑔 of π‘₯ plus 𝑐 is simply a vertical translation of the graph of 𝑦 equals 𝑔 of π‘₯ by 𝑐 units.

Recalling that the first derivative of a function gives its slope at that point, we see that the slope of the two graphs is identical for each π‘₯-value. One is just positioned vertically above the other. Therefore, we see again that adding a constant to a function makes no difference to its derivative. And so, the derivative of a constant term is just zero.

Finally, we consider the derivative of the sum or difference of two or more functions. By returning to first principles, we could prove that the derivative of the sum or difference of two functions is equal to the sum or difference of the derivatives of those functions. The derivative with respect to π‘₯ of 𝑓 of π‘₯ plus or minus 𝑔 of π‘₯ is the derivative with respect to π‘₯ of 𝑓 of π‘₯ plus or minus the derivative with respect to π‘₯ of 𝑔 of π‘₯. As the polynomial is just the sum of power terms, which we now know how to differentiate, we can find the derivative of polynomial functions by combining these rules, without the need to return to first principles. Let’s now consider some examples of this.

Find d𝑦 by dπ‘₯, given that 𝑦 equals 22π‘₯ to the fourth power.

The function we’ve been asked to differentiate is just a general power term. So we can answer this question using the power rule of differentiation. This tells us that for real constants π‘Ž and 𝑛, the derivative with respect to π‘₯ of π‘Žπ‘₯ to the 𝑛th power is π‘Žπ‘›π‘₯ to the power of 𝑛 minus one. We multiply by the original exponent and then reduce the exponent by one. In this question, the value of π‘Ž, which is the coefficient, is 22. And the value of 𝑛, which is the exponent, is four. So applying the power rule then, we have that d𝑦 by dπ‘₯ is equal to π‘Ž, that’s 22, multiplied by 𝑛, that’s four, multiplied by π‘₯ to the power of four minus one.

In practical terms, remember, what we’ve done is multiply the original exponent and then reduce the exponent by one. We can simplify the coefficient; 22 multiplied by four is 88. And then simplify the exponent; four minus one is three. So by applying the power rule of differentiation, we’ve found that if 𝑦 equals 22π‘₯ to the fourth power, d𝑦 by dπ‘₯ equals 88π‘₯ cubed.

Find the first derivative of the function 𝑦 equals two π‘₯ multiplied by nine π‘₯ squared minus three π‘₯ plus 10π‘₯.

To begin, let’s simplify our function 𝑦 by distributing the parentheses. We have two π‘₯ multiplied by nine π‘₯ squared, which gives 18π‘₯ cubed. And then two π‘₯ multiplied by negative three π‘₯, which gives negative six π‘₯ squared. So our function 𝑦 is equal to 18π‘₯ cubed minus six π‘₯ squared plus 10π‘₯. We then see that our function 𝑦 is just the sum of power terms of π‘₯. It is a polynomial. We can therefore find its derivative using the power rule. Which tells us that the derivative with respect to π‘₯ of π‘Žπ‘₯ to the 𝑛th power for real constants π‘Ž and 𝑛 is equal to π‘Žπ‘›π‘₯ to the power of 𝑛 minus one. We multiply by the exponent and then reduce the exponent by one.

We also need to remember that in order to find the derivative of the sum or difference of functions, we can just take the sum or difference of their derivatives. So we can differentiate each term separately and then add or subtract. Let’s differentiate term by term then. The derivative of 18π‘₯ cubed is 18 multiplied by three π‘₯ squared. We’ve multiplied by the exponent of three and then reduced the exponent by one. The derivative of negative six π‘₯ squared is negative six multiplied by two π‘₯. And to find the derivative of 10π‘₯, we may find it helpful to think of this as 10π‘₯ to the power of one. The derivative is therefore 10 multiplied by one multiplied by π‘₯ to the power of zero.

Remember, though, that π‘₯ to the power of zero is just one. So the derivative of 10π‘₯ is just 10. Simplifying the coefficients, and we have that d𝑦 by dπ‘₯ is equal to 54π‘₯ squared minus 12π‘₯ plus 10.

In our next example, we’ll see how to find the derivative of a power term with a negative exponent.

Find d𝑦 by dπ‘₯, given that 𝑦 equals negative 43 over π‘₯ to the eighth power.

The first step in answering this question is to recall that a reciprocal can be rewritten using a negative exponent. One over π‘₯ to the 𝑛th power is equal to π‘₯ to the power of negative 𝑛. So we can rewrite our function 𝑦 as 𝑦 equals negative 43π‘₯ to the power of negative eight. And now, we see that it consists of a general power term whose derivative we can find using the power rule of differentiation. This tells us that for real constants π‘Ž and 𝑛, the derivative with respect to π‘₯ of π‘Žπ‘₯ to the 𝑛th power is equal to π‘Žπ‘›π‘₯ to the power of 𝑛 minus one.

So applying the power rule, we have that d𝑦 by dπ‘₯ is equal to negative 43 multiplied by negative eight π‘₯ to the power of negative nine. We do need to be really careful with the exponent here. That value of negative nine comes from reducing the original exponent of negative eight by one. Remember, we are subtracting one. But because it is a negative exponent, a common mistake is to change the exponent to negative seven. But that would actually be increasing it by one. Make sure you watch out for this when differentiating terms with negative exponents.

We can then simplify the coefficient. Negative 43 multiplied by negative eight gives 344. And then finally, rewrite π‘₯ to the power of negative nine as a reciprocal. We’ve found then that if 𝑦 is equal to negative 43 over π‘₯ to the eighth power, d𝑦 by dπ‘₯ is equal to 344 over π‘₯ to the ninth power.

In our final example, we’ll see how to find the derivative of an expression involving a square root.

Find d𝑦 by dπ‘₯ if 𝑦 equals six root π‘₯ over seven.

In order to answer this question, we need to think about the alternative ways that we can express a square root. We know from our laws of exponents that the 𝑛th root of π‘₯ can be rewritten as π‘₯ to the power of one over 𝑛. Although we don’t write the small two for a square root, that is our value of 𝑛. So we can rewrite the square root of π‘₯ as π‘₯ to the power of one-half. Our expression for 𝑦 can therefore be rewritten as six-sevenths π‘₯ to the power of one-half. And we see that we have a general power term. We can differentiate this using the power rule of differentiation which tells us that for real constants π‘Ž and 𝑛, the derivative with respect to π‘₯ of π‘Žπ‘₯ to the 𝑛th power is equal to π‘Žπ‘›π‘₯ to the power of 𝑛 minus one.

So here we go then. We, first of all, multiplied by the exponent 𝑛; that’s one-half. And then, we’ve reduced the exponent by one, giving d𝑦 by dπ‘₯ equals six-sevenths multiplied by one-half multiplied by π‘₯ to the power of one-half minus one. We can simplify the coefficient by cancelling a shared factor of two. And then one-half minus one is equal to negative one-half. We then recall another of our laws of exponents which is that a negative exponent defines a reciprocal. So π‘₯ to the power of negative one-half is one over π‘₯ to the power of a half or one over the square root of π‘₯. We’ve found then that d𝑦 by dπ‘₯ is equal to three over seven root π‘₯.

Let’s now summarize what we’ve seen in this video. In its most simple form, the power rule of differentiation tells us that for a real exponent 𝑛, the derivative with respect to π‘₯ of π‘₯ to the 𝑛th power is 𝑛π‘₯ to the power of 𝑛 minus one. More generally, if we have a coefficient of π‘Ž, the derivative with respect to π‘₯ of π‘Žπ‘₯ to the 𝑛th power is π‘Žπ‘›π‘₯ to the power of 𝑛 minus one. We also saw that an application of this is that the derivative with respect to π‘₯ of any constant 𝑐 is just zero.

Finally, we saw that the derivative of a sum or difference of terms or functions is just the sum or difference of their individual derivatives. We saw how to apply these rules to differentiating powers of π‘₯, including with fractional or negative exponents. And we saw how to differentiate polynomials without returning to first principles. In future, we’ll also be able to apply these rules to differentiate a wider class of functions.

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