### Video Transcript

In this video, weβll learn how to
find second- and higher-order derivatives of functions, including those that require
a variety of rules for differentiation. At this stage, you should feel
comfortable differentiating polynomial and trigonometric functions and applying the
chain rule, product rule, and quotient rule. Weβre going to use these skills to
discover how to find second- and higher-order derivatives.

Because the derivative of a
function π¦ equals π of π₯ is itself a function in π₯, this means we can take the
derivative of the derivative, or the derivative of π prime of π₯. This is known as taking the second
derivative of the function. And itβs denoted as π prime prime
of π₯, or d two π¦ by dπ₯ two, or d two π¦ by dπ₯ squared, using Leibnizβs
notation.

This idea can be continued to find
the third, fourth, fifth, and successive derivatives of the original function. After the third derivative, we
donβt continue to use the prime notation as it begins to get a little
cumbersome. And instead, we denote the πth
derivative as shown. Generally, we perform this process
in a sequential manner, differentiating each function in π₯ such that the πth
derivative of the function is equal to the derivative of the π minus oneth
derivative of the function.

Itβs also useful to note though
that there are some cases in which we can derive a general formula for an arbitrary
πth-order derivative. And weβll consider these later
on. There are also a number of
applications of second- and higher-order derivatives, though we wonβt be considering
these in this video, just focusing on the processes as required. Letβs consider first a simple
example that involves finding the second derivative of a polynomial function.

Given that π¦ is equal to six
π₯ to the power of five plus three π₯ squared minus seven π₯ plus six, determine
the second derivative of π¦ with respect to π₯.

Here weβve been given a
function in π₯. And weβre being asked to find d
two π¦ by dπ₯ squared. To do this, weβll differentiate
once to find dπ¦ by dπ₯ then differentiate that once again to find the second
derivative. We recall that we can
differentiate a function of the form ππ₯ to the power of π with respect to π₯
for some constant rational number π, which is not equal to zero and some
constant π. And we get ππ times π₯ to the
power of π minus one. In other words, we borrow the
power of π₯ and we make it the coefficient of the derivative. And then, we subtract one from
the power.

In this special case where π
is equal to zero, we actually have a constant. Letβs call that π. And the derivative of a
constant is zero. This is going to help us find
the first derivative of our function. The derivative of six π₯ to the
power of five is going to be five times six π₯. And then, we subtract one from
the power. Five minus one is four. Thatβs 30π₯ to the fourth
power. Weβll repeat this to
differentiate three π₯ squared. Itβs going to be two times
three π₯. And then, we subtract one from
the power. Two minus one is one. So, the derivative of three π₯
squared with respect to π₯ is six π₯.

The derivative of negative
seven π₯ is one times negative seven π₯ to the power of zero. Well, thatβs just negative
seven. And, of course, six is a
constant, so the derivative of six is zero. dπ¦ by dπ₯ then, the first
derivative of our equation, is 30π₯ to the power of four plus six π₯ minus
seven. Weβll differentiate each part
of this expression once more to find the second derivative.

Weβll do it piece-by-piece. The derivative of 30π₯ to the
power of four is four times 30π₯ to the power of three. The derivative of six π₯ is
six. And the derivative of negative
seven is zero. Simplifying fully, and we see
that the second derivative of π¦ with respect to π₯ is 120π₯ cubed plus six.

In this example, we saw how we
could apply repeated differentiation to help us find the second derivative of a
polynomial function.

Next, weβll look at how we can
evaluate the second derivative of a slightly more complicated example at a given
point.

Determine the value of the
second derivative of the function π¦ equals 12π₯ minus eight over π₯ at one,
four.

We have an equation for π¦ in
terms of π₯. And weβre being asked to find
the value of the second derivative at the point with Cartesian coordinates one,
four. Weβll begin then by simply
finding an expression for the second derivative. To do this, weβll differentiate
our function once to find dπ¦ by dπ₯ then differentiate with respect to π₯ once
more. It might help, before we do, to
write π¦ as 12π₯ minus eight π₯ to the negative one. Then, we differentiate as
normal.

The derivative of 12π₯ with
respect to π₯ is simply 12. And the derivative of negative
eight π₯ to the power of negative one is negative negative one times eight π₯ to
the power of negative two. And be really careful. A common mistake here is to
spot the negative one and think that when we subtract one, we get to zero. When we simplify this, we see
that the first derivative is 12 plus eight π₯ to the negative two. Letβs repeat this process to
find the second derivative.

The derivative of 12 is
zero. And then, when we differentiate
eight π₯ to the negative two with respect to π₯, we get negative two times eight
π₯ to the negative three. Thatβs negative 16π₯ to the
negative three. And, of course, we can change
this back to negative 16 over π₯ cubed if we like.

We need to determine the value
of the second derivative at one, four. This is a Cartesian
coordinate. It has an π₯-value of one and a
π¦-value of four. So, weβre going to substitute
π₯ is equal to one into our equation for the second derivative. This gives us negative 16 over
one cubed, which is negative 16.

In our next example, weβre going to
learn how we can apply the standard rules for differentiation to help us find second
derivatives.

Given that π¦ is equal to three
π₯ squared minus five over two π₯ squared plus seven, determine the second
derivative of π¦ with respect to π₯.

Here we have a quotient. Itβs the result of dividing one
function by another function. We can, therefore, use the
quotient rule to help us find the first derivative. This says that for two
differentiable functions π’ and π£, the derivative of π’ over π£ with respect to
π₯ is equal to π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£
squared.

Since π’ is the numerator,
weβre going to let π’ be equal to three π₯ squared minus five. And we can see that π£ must be
equal to two π₯ squared plus seven. To be able to use the quotient
rule, weβre going to differentiate each of these with respect to π₯. When we differentiate π’ with
respect to π₯, we get six π₯. And dπ£ by dπ₯ is equal to four
π₯. π£ times dπ’ by dπ₯ is two π₯
squared plus seven times six π₯. We then subtract the product of
π’ and dπ£ by dπ₯. Thatβs three π₯ squared minus
five times four π₯. And, of course, thatβs all over
π£ squared. Thatβs two π₯ squared plus
seven squared.

Distributing the parentheses on
the top of our fraction, and we get 12π₯ cubed plus 42π₯ minus 12π₯ cubed plus
20π₯. And then, weβll keep the
denominator as is. And we see that the derivative
of π¦ with respect π₯ is 62π₯ over two π₯ squared plus seven squared.

To find the second derivative,
weβre going to need to differentiate this again. Letβs clear some space. Once again, weβre looking to
differentiate a quotient, so weβre going to use the quotient rule. This time, we let π’ be equal
to 62π₯ and π£ be equal to two π₯ squared plus seven squared. dπ’ by dπ₯ is
fairly straightforward. Itβs 62. But what about dπ£ by dπ₯?

Well, we can use a special case
of the chain rule called the general power rule. This says that if π is some
function of π₯ and π is some constant rational not equal to zero, then the
derivative of π to the power of π with respect to π₯ is π times π to the
power of π minus one times dπ by dπ₯. This means that dπ£ by dπ₯ is
two times this function in π₯, which is two π₯ squared plus seven to the power
of one, times the derivative of two π₯ squared plus seven with respect π₯, which
is four π₯.

And when we simplify this, we
see that dπ£ by dπ₯ is eight π₯ times two π₯ squared plus seven. This time, π£ times dπ’ by dπ₯
is two π₯ squared plus seven squared times 62 minus π’ times dπ£ by dπ₯ all over
π£ squared. The key here is to spot that
the denominator of our fraction becomes two π₯ squared plus seven to the power
of four. And this means we can divide
through by a common factor of two π₯ squared plus seven. And that leaves us with two π₯
squared plus seven cubed on the denominator and 62 times two π₯ squared plus
seven minus 62π₯ times eight π₯ on the top.

We can take out a factor of 62
on the numerator. And then, when we subtract
eight π₯ times π₯ from two π₯ squared, we get negative six π₯ squared. And the second derivative of π¦
with respect to π₯ is 62 times seven minus six squared over two π₯ squared plus
seven cubed.

Now that weβve seen how to find the
second derivative, weβre going to look at using higher-order derivatives to solve
problems.

Given that π¦ is equal to ππ₯
cubed plus ππ₯ squared, the third derivative of π¦ is negative 18, and the
second derivative of π¦ with respect to π₯ evaluated that π₯ equals two is
negative 14, find π and π.

Here, we have an equation for
π¦ in terms of π₯ and some information about the second derivative and the third
derivative, denoted as π¦ prime prime prime. To answer this question, letβs
begin just by finding an equation for the second and third derivatives of π¦
with respect to π₯.

Differentiating π¦ with respect
to π₯, and we get three ππ₯ squared plus two ππ₯. To find the second derivative,
weβll differentiate the equation for the second derivative. Thatβs two times three ππ₯
plus two π. That simplifies to six ππ₯
plus two π. Once again, to find the third
derivative, we differentiate the second derivative with respect to π₯. Since two π is a constant, the
third derivative is six π.

Weβre told that the second
derivative evaluated at π₯ is equal to two is negative 14. So, letβs substitute π₯ is
equal to two into our equation for the second derivative and set it equal to
negative 14. Thatβs six times two plus two
π equals negative 14, or 12π plus two π is negative 14. Similarly, weβre told that the
third derivative is equal to negative 18. So, we can say that six π must
be equal to negative 18.

Notice, that this latter
equation has a single variable, so we can solve as normal. We can divide both sides of
this equation by six. And when we do, we see that π
is equal to negative three. We can take this value and
substitute it into the equation we formed using the second derivative. That gives us 12 multiplied by
negative three plus two π equals negative 14. 12 multiplied by negative three
is negative 36. We add 36 to both sides of our
equation to get two π equals 22. And we divide through by two to
get π equals 11. π equals negative three and π
equals 11.

In our final example, weβll
consider a case in which we can derive a general formula for an arbitrary πth-order
derivative.

Find the 51st derivative of sin
of π₯ with respect to π₯ by finding the first few derivatives and observing the
pattern that occurs.

Letβs begin then by finding the
first few derivatives of sin of π₯ with respect to π₯. We can quote the standard
result that d by dπ₯ of sin of π₯ is cos of π₯. This means to find the second
derivative of sin of π₯, we need to differentiate cos of π₯ with respect to
π₯.

Here, we can quote another
standard result. The derivative of cos of π₯
with respect to π₯ is negative sin of π₯. So, the second derivative of
sin of π₯ with respect to π₯ is negative sin of π₯. Similarly, the third derivative
is going to be found by differentiating negative sin π₯ with respect to π₯. And we can use the constant
multiple rule here to take the constant of negative one outside the derivative
and concentrate on differentiating sin of π₯.

Weβve already seen that the
derivative of sin of π₯ with respect to π₯ is cos of π₯. So, this means that the third
derivative of sin of π₯ with respect to π₯ is negative cos of π₯. The fourth derivative of sin of
π₯ is going to be the derivative of negative cos of π₯ with respect to π₯. Once again, weβll use the
constant multiple rule here and take the constant of negative one outside the
derivative and concentrate on differentiating cos of π₯, which we now know to be
negative sin of π₯.

So, the fourth derivative is
negative negative sin of π₯, which is positive sin of π₯. And we donβt actually need to
do anymore. We can see that we have a
cycle. The fifth derivative of sin of
π₯ is going to be cos of π₯. And the sixth derivative will
go back to negative sin of π₯, and so on. So, whatβs the general
rule?

Well, we can say that for
integer values of π, the four πth derivative of sin of π₯ is sin of π₯. The four πth plus oneth
derivative of sin of π₯ is cos of π₯. The four π plus twoth
derivative of sin of π₯ is negative sin of π₯. And the four π plus threeth
derivative of sin of π₯ is negative cos of π₯.

Weβre trying to find the 51st
derivative. And we can write 51 as four
times 12 plus three. So, this means that the 51st
derivative of sin of π₯ will be the same as the four π plus threeth
derivative. Itβs negative cos of π₯.

Itβs useful to know that since
the derivatives of sin and cos are so closely related, we can also derive a
general formula for the πth derivative of cos of π₯. The four πth derivative of cos
of π₯ is cos of π₯. The four π plus oneth
derivative of cos of π₯ is negative sin of π₯. The four π plus twoth
derivative of cos of π₯ is negative cos of π₯. And the four π plus threeth
derivative of cos of π₯ is sin of π₯.

In this video, weβve seen that we
can use the standard rules for differentiation to find second- and higher-order
derivatives. We saw that we generally perform
this process in a sequential manner, though there are occasions where we can use
patterns to generate rules for higher-order derivatives.