### 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.