Video Transcript
In this video, we’ll 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. We’re going to use these skills to
discover how to find second- and higher-order derivatives.
Because the derivative of a
function 𝑦 equals 𝑓 of 𝑥 is itself a function in 𝑥, this means we can take the
derivative of the derivative, or the derivative of 𝑓 prime of 𝑥. This is known as taking the second
derivative of the function. And it’s denoted as 𝑓 prime prime
of 𝑥, or d two 𝑦 by d𝑥 two, or d two 𝑦 by d𝑥 squared, using Leibniz’s
notation.
This idea can be continued to find
the third, fourth, fifth, and successive derivatives of the original function. After the third derivative, we
don’t continue to use the prime notation as it begins to get a little
cumbersome. And instead, we denote the 𝑛th
derivative as shown. Generally, we perform this process
in a sequential manner, differentiating each function in 𝑥 such that the 𝑛th
derivative of the function is equal to the derivative of the 𝑛 minus oneth
derivative of the function.
It’s also useful to note though
that there are some cases in which we can derive a general formula for an arbitrary
𝑛th-order derivative. And we’ll consider these later
on. There are also a number of
applications of second- and higher-order derivatives, though we won’t be considering
these in this video, just focusing on the processes as required. Let’s consider first a simple
example that involves finding the second derivative of a polynomial function.
Given that 𝑦 is equal to six
𝑥 to the power of five plus three 𝑥 squared minus seven 𝑥 plus six, determine
the second derivative of 𝑦 with respect to 𝑥.
Here we’ve been given a
function in 𝑥. And we’re being asked to find d
two 𝑦 by d𝑥 squared. To do this, we’ll differentiate
once to find d𝑦 by d𝑥 then differentiate that once again to find the second
derivative. We recall that we can
differentiate a function of the form 𝑎𝑥 to the power of 𝑛 with respect to 𝑥
for some constant rational number 𝑛, which is not equal to zero and some
constant 𝑎. And we get 𝑛𝑎 times 𝑥 to the
power of 𝑛 minus one. In other words, we borrow the
power of 𝑥 and we make it the coefficient of the derivative. And then, we subtract one from
the power.
In this special case where 𝑛
is equal to zero, we actually have a constant. Let’s call that 𝑏. 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 𝑥 to the
power of five is going to be five times six 𝑥. And then, we subtract one from
the power. Five minus one is four. That’s 30𝑥 to the fourth
power. We’ll repeat this to
differentiate three 𝑥 squared. It’s going to be two times
three 𝑥. And then, we subtract one from
the power. Two minus one is one. So, the derivative of three 𝑥
squared with respect to 𝑥 is six 𝑥.
The derivative of negative
seven 𝑥 is one times negative seven 𝑥 to the power of zero. Well, that’s just negative
seven. And, of course, six is a
constant, so the derivative of six is zero. d𝑦 by d𝑥 then, the first
derivative of our equation, is 30𝑥 to the power of four plus six 𝑥 minus
seven. We’ll differentiate each part
of this expression once more to find the second derivative.
We’ll do it piece-by-piece. The derivative of 30𝑥 to the
power of four is four times 30𝑥 to the power of three. The derivative of six 𝑥 is
six. And the derivative of negative
seven is zero. Simplifying fully, and we see
that the second derivative of 𝑦 with respect to 𝑥 is 120𝑥 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, we’ll 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 𝑦 equals 12𝑥 minus eight over 𝑥 at one,
four.
We have an equation for 𝑦 in
terms of 𝑥. And we’re being asked to find
the value of the second derivative at the point with Cartesian coordinates one,
four. We’ll begin then by simply
finding an expression for the second derivative. To do this, we’ll differentiate
our function once to find d𝑦 by d𝑥 then differentiate with respect to 𝑥 once
more. It might help, before we do, to
write 𝑦 as 12𝑥 minus eight 𝑥 to the negative one. Then, we differentiate as
normal.
The derivative of 12𝑥 with
respect to 𝑥 is simply 12. And the derivative of negative
eight 𝑥 to the power of negative one is negative negative one times eight 𝑥 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 𝑥 to the negative two. Let’s repeat this process to
find the second derivative.
The derivative of 12 is
zero. And then, when we differentiate
eight 𝑥 to the negative two with respect to 𝑥, we get negative two times eight
𝑥 to the negative three. That’s negative 16𝑥 to the
negative three. And, of course, we can change
this back to negative 16 over 𝑥 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 𝑥-value of one and a
𝑦-value of four. So, we’re going to substitute
𝑥 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, we’re going to
learn how we can apply the standard rules for differentiation to help us find second
derivatives.
Given that 𝑦 is equal to three
𝑥 squared minus five over two 𝑥 squared plus seven, determine the second
derivative of 𝑦 with respect to 𝑥.
Here we have a quotient. It’s 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 𝑢 and 𝑣, the derivative of 𝑢 over 𝑣 with respect to
𝑥 is equal to 𝑣 times d𝑢 by d𝑥 minus 𝑢 times d𝑣 by d𝑥 all over 𝑣
squared.
Since 𝑢 is the numerator,
we’re going to let 𝑢 be equal to three 𝑥 squared minus five. And we can see that 𝑣 must be
equal to two 𝑥 squared plus seven. To be able to use the quotient
rule, we’re going to differentiate each of these with respect to 𝑥. When we differentiate 𝑢 with
respect to 𝑥, we get six 𝑥. And d𝑣 by d𝑥 is equal to four
𝑥. 𝑣 times d𝑢 by d𝑥 is two 𝑥
squared plus seven times six 𝑥. We then subtract the product of
𝑢 and d𝑣 by d𝑥. That’s three 𝑥 squared minus
five times four 𝑥. And, of course, that’s all over
𝑣 squared. That’s two 𝑥 squared plus
seven squared.
Distributing the parentheses on
the top of our fraction, and we get 12𝑥 cubed plus 42𝑥 minus 12𝑥 cubed plus
20𝑥. And then, we’ll keep the
denominator as is. And we see that the derivative
of 𝑦 with respect 𝑥 is 62𝑥 over two 𝑥 squared plus seven squared.
To find the second derivative,
we’re going to need to differentiate this again. Let’s clear some space. Once again, we’re looking to
differentiate a quotient, so we’re going to use the quotient rule. This time, we let 𝑢 be equal
to 62𝑥 and 𝑣 be equal to two 𝑥 squared plus seven squared. d𝑢 by d𝑥 is
fairly straightforward. It’s 62. But what about d𝑣 by d𝑥?
Well, we can use a special case
of the chain rule called the general power rule. This says that if 𝑝 is some
function of 𝑥 and 𝑛 is some constant rational not equal to zero, then the
derivative of 𝑝 to the power of 𝑛 with respect to 𝑥 is 𝑛 times 𝑝 to the
power of 𝑛 minus one times d𝑝 by d𝑥. This means that d𝑣 by d𝑥 is
two times this function in 𝑥, which is two 𝑥 squared plus seven to the power
of one, times the derivative of two 𝑥 squared plus seven with respect 𝑥, which
is four 𝑥.
And when we simplify this, we
see that d𝑣 by d𝑥 is eight 𝑥 times two 𝑥 squared plus seven. This time, 𝑣 times d𝑢 by d𝑥
is two 𝑥 squared plus seven squared times 62 minus 𝑢 times d𝑣 by d𝑥 all over
𝑣 squared. The key here is to spot that
the denominator of our fraction becomes two 𝑥 squared plus seven to the power
of four. And this means we can divide
through by a common factor of two 𝑥 squared plus seven. And that leaves us with two 𝑥
squared plus seven cubed on the denominator and 62 times two 𝑥 squared plus
seven minus 62𝑥 times eight 𝑥 on the top.
We can take out a factor of 62
on the numerator. And then, when we subtract
eight 𝑥 times 𝑥 from two 𝑥 squared, we get negative six 𝑥 squared. And the second derivative of 𝑦
with respect to 𝑥 is 62 times seven minus six squared over two 𝑥 squared plus
seven cubed.
Now that we’ve seen how to find the
second derivative, we’re going to look at using higher-order derivatives to solve
problems.
Given that 𝑦 is equal to 𝑎𝑥
cubed plus 𝑏𝑥 squared, the third derivative of 𝑦 is negative 18, and the
second derivative of 𝑦 with respect to 𝑥 evaluated that 𝑥 equals two is
negative 14, find 𝑎 and 𝑏.
Here, we have an equation for
𝑦 in terms of 𝑥 and some information about the second derivative and the third
derivative, denoted as 𝑦 prime prime prime. To answer this question, let’s
begin just by finding an equation for the second and third derivatives of 𝑦
with respect to 𝑥.
Differentiating 𝑦 with respect
to 𝑥, and we get three 𝑎𝑥 squared plus two 𝑏𝑥. To find the second derivative,
we’ll differentiate the equation for the second derivative. That’s two times three 𝑎𝑥
plus two 𝑏. That simplifies to six 𝑎𝑥
plus two 𝑏. Once again, to find the third
derivative, we differentiate the second derivative with respect to 𝑥. Since two 𝑏 is a constant, the
third derivative is six 𝑎.
We’re told that the second
derivative evaluated at 𝑥 is equal to two is negative 14. So, let’s substitute 𝑥 is
equal to two into our equation for the second derivative and set it equal to
negative 14. That’s six times two plus two
𝑏 equals negative 14, or 12𝑎 plus two 𝑏 is negative 14. Similarly, we’re told that the
third derivative is equal to negative 18. So, we can say that six 𝑎 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 𝑎
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 𝑏 equals negative 14. 12 multiplied by negative three
is negative 36. We add 36 to both sides of our
equation to get two 𝑏 equals 22. And we divide through by two to
get 𝑏 equals 11. 𝑎 equals negative three and 𝑏
equals 11.
In our final example, we’ll
consider a case in which we can derive a general formula for an arbitrary 𝑛th-order
derivative.
Find the 51st derivative of sin
of 𝑥 with respect to 𝑥 by finding the first few derivatives and observing the
pattern that occurs.
Let’s begin then by finding the
first few derivatives of sin of 𝑥 with respect to 𝑥. We can quote the standard
result that d by d𝑥 of sin of 𝑥 is cos of 𝑥. This means to find the second
derivative of sin of 𝑥, we need to differentiate cos of 𝑥 with respect to
𝑥.
Here, we can quote another
standard result. The derivative of cos of 𝑥
with respect to 𝑥 is negative sin of 𝑥. So, the second derivative of
sin of 𝑥 with respect to 𝑥 is negative sin of 𝑥. Similarly, the third derivative
is going to be found by differentiating negative sin 𝑥 with respect to 𝑥. 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 𝑥.
We’ve already seen that the
derivative of sin of 𝑥 with respect to 𝑥 is cos of 𝑥. So, this means that the third
derivative of sin of 𝑥 with respect to 𝑥 is negative cos of 𝑥. The fourth derivative of sin of
𝑥 is going to be the derivative of negative cos of 𝑥 with respect to 𝑥. Once again, we’ll use the
constant multiple rule here and take the constant of negative one outside the
derivative and concentrate on differentiating cos of 𝑥, which we now know to be
negative sin of 𝑥.
So, the fourth derivative is
negative negative sin of 𝑥, which is positive sin of 𝑥. And we don’t actually need to
do anymore. We can see that we have a
cycle. The fifth derivative of sin of
𝑥 is going to be cos of 𝑥. And the sixth derivative will
go back to negative sin of 𝑥, and so on. So, what’s the general
rule?
Well, we can say that for
integer values of 𝑘, the four 𝑘th derivative of sin of 𝑥 is sin of 𝑥. The four 𝑘th plus oneth
derivative of sin of 𝑥 is cos of 𝑥. The four 𝑘 plus twoth
derivative of sin of 𝑥 is negative sin of 𝑥. And the four 𝑘 plus threeth
derivative of sin of 𝑥 is negative cos of 𝑥.
We’re 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 𝑥 will be the same as the four 𝑘 plus threeth
derivative. It’s negative cos of 𝑥.
It’s useful to know that since
the derivatives of sin and cos are so closely related, we can also derive a
general formula for the 𝑛th derivative of cos of 𝑥. The four 𝑘th derivative of cos
of 𝑥 is cos of 𝑥. The four 𝑘 plus oneth
derivative of cos of 𝑥 is negative sin of 𝑥. The four 𝑘 plus twoth
derivative of cos of 𝑥 is negative cos of 𝑥. And the four 𝑘 plus threeth
derivative of cos of 𝑥 is sin of 𝑥.
In this video, we’ve 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.
