### 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. Weβll 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. Weβll then apply these rules to a
series of examples.

Letβs begin by recalling the form
or definition of the derivative of a function. For the function π of π₯, its
derivative, denoted as π prime of π₯, is given by the limit as β tends to zero of
π of π₯ plus β minus π of π₯ over β at the points where this limit exists. Where does this definition come
from? Well, letβs consider the graph of a
function π¦ equals π 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, 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 π¦ over
change in π₯. The change in π¦ will be the
function evaluated at π₯ plus β minus the function evaluated at π₯. Thatβs π of π₯ plus β minus π of
π₯. And the horizontal change will be
π₯ plus β minus π₯. Of course, the π₯s in the
denominator cancel. And weβre just left with π of π₯
plus β minus π of π₯ over β. Remember, though, that this is the
slope of this short chord, not the slope of the curve itself. Although if β 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 so that
these two points get closer together. As β tends to zero, this sequence
of chords will tend to the tangent to the graph of π at point π₯. The slope of the tangent then β
which, remember, is the slope of the curve and its first derivative, π prime of π₯
β is the limit as β tends to zero of this quantity, π of π₯ plus β minus π of π₯
over β. Which is what we wrote down earlier
in our formal definition.

So thatβs 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, letβs use this
definition to find an expression for the derivative of the function π of π₯ equals
π₯ squared, which is called differentiating from first principles. And then, weβll use this to see if
we can generalize.

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

So using differentiation from first
principles, weβve 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, weβd 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 weβve 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 πth power is equal to π multiplied by π₯ to the
power of π minus one. We reduce the exponent by one and
multiply it by the original exponent.

Now, itβs important to note that we
have only demonstrated this general rule. We havenβt proven it. To do so, at least for positive
integer values of π, would require an application of the binomial theorem. But itβs beyond the scope of this
video to look at that here. This rule actually works for any
real value of π. So that includes positive,
negative, and fractional exponents.

So thatβs great if the coefficient
of our term is just one. But what about if we have a general
coefficient π? So we have a constant multiplied by
π₯ to some real exponent π. Now, we could return to first
principles, perhaps using the example function π 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, weβd find that π prime of π₯
is equal to eight π₯. Now, how does this link to what
weβve 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 weβve 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 multiplied by
π₯ to the πth power for real constants π and π is equal to π multiplied by ππ₯
to the power of π 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 π¦ equals four π₯
squared is a vertical stretch of the graph of π¦ 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 π¦ equals four π₯ squared compared to what it was on the graph
of π¦ equals π₯ squared. And so, we have an extra factor of
four in our derivative or, more generally, an extra factor of π.

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 π 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 π 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
π of π₯ equals seven and, in fact, the derivative of any constant π with respect
to π₯ is just zero.

We can, again, see the practical
interpretation of this if we consider the graph of π of π₯ equals seven. Itβs a horizontal line through
seven on the vertical axis. If we recall that a derivative
gives the rate of change of π¦ with respect to π₯, then we see that as π¦ isnβt
changing, its rate of change will be zero. In the same way, if we were to
consider the graphs of π¦ equals π of π₯ and π¦ equals π of π₯ plus π for some
function π of π₯ and some constant π. We know that the graph of π¦ equals
π of π₯ plus π is simply a vertical translation of the graph of π¦ equals π of π₯
by π 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 π of π₯ plus or minus π of π₯ is the derivative with respect to π₯ of π of π₯
plus or minus the derivative with respect to π₯ of π 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. Letβs now consider some examples of
this.

Find dπ¦ by dπ₯, given that π¦
equals 22π₯ to the fourth power.

The function weβve 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 and π, the derivative with respect to π₯ of ππ₯ to the πth power is
πππ₯ to the power of π minus one. We multiply by the original
exponent and then reduce the exponent by one. In this question, the value of π,
which is the coefficient, is 22. And the value of π, which is the
exponent, is four. So applying the power rule then, we
have that dπ¦ by dπ₯ is equal to π, thatβs 22, multiplied by π, thatβs four,
multiplied by π₯ to the power of four minus one.

In practical terms, remember, what
weβve 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, weβve found that if π¦ equals 22π₯ to the fourth power, dπ¦ by dπ₯
equals 88π₯ cubed.

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

To begin, letβs simplify our
function π¦ 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 π¦ is equal to 18π₯
cubed minus six π₯ squared plus 10π₯. We then see that our function π¦ 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 ππ₯ to the πth power for real constants π and π is equal
to πππ₯ to the power of π 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. Letβs differentiate term by term
then. The derivative of 18π₯ cubed is 18
multiplied by three π₯ squared. Weβve 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 dπ¦ by dπ₯ is equal to 54π₯ squared minus 12π₯ plus 10.

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

Find dπ¦ by dπ₯, given that π¦
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 πth power is
equal to π₯ to the power of negative π. So we can rewrite our function π¦
as π¦ 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 and π, the derivative with respect to π₯ of ππ₯ to the πth power is
equal to πππ₯ to the power of π minus one.

So applying the power rule, we have
that dπ¦ 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. Weβve found then that if π¦ is
equal to negative 43 over π₯ to the eighth power, dπ¦ by dπ₯ is equal to 344 over π₯
to the ninth power.

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

Find dπ¦ by dπ₯ if π¦ 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 πth root of π₯ can be rewritten as π₯ to the power of one over π. Although we donβt write the small
two for a square root, that is our value of π. So we can rewrite the square root
of π₯ as π₯ to the power of one-half. Our expression for π¦ 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 and π, the
derivative with respect to π₯ of ππ₯ to the πth power is equal to πππ₯ to the
power of π minus one.

So here we go then. We, first of all, multiplied by the
exponent π; thatβs one-half. And then, weβve reduced the
exponent by one, giving dπ¦ 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₯. Weβve found then that dπ¦ by dπ₯ is
equal to three over seven root π₯.

Letβs now summarize what weβve seen
in this video. In its most simple form, the power
rule of differentiation tells us that for a real exponent π, the derivative with
respect to π₯ of π₯ to the πth power is ππ₯ to the power of π minus one. More generally, if we have a
coefficient of π, the derivative with respect to π₯ of ππ₯ to the πth power is
πππ₯ to the power of π minus one. We also saw that an application of
this is that the derivative with respect to π₯ of any constant π 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, weβll also be able to
apply these rules to differentiate a wider class of functions.