### Video Transcript

In this lesson, weβll learn how to find the derivative of trigonometric functions, focusing on the cotangent, secant, and cosecant functions. Weβll begin by recapping the rules that we need to find these derivatives before completing the differentiation and then considering how these standard results can help us to find the derivatives of more complex functions.

Letβs begin by recapping the derivative of the trigonometric functions sine, cosine, and tangent. The derivative of sin of ππ₯ is π cos of ππ₯. The derivative of cos ππ₯ is negative π sin ππ₯. And the derivative of tan ππ₯ is π sec squared ππ₯. Weβre also going to need to use the product rule. And this is if we have two differentiable functions, π’ and π£, the derivative of their product, π’ times π£, is π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. Similarly, weβll use the quotient rule. And this time, we find the derivative of the quotient of two differentiable functions, π’ and π£, by finding π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£ squared. Weβll be referring to each of these throughout the video. So letβs look at first example.

If π¦ is equal to negative two sec of two π₯, determine the rate of change of π¦ when π₯ is equal to 11π over six.

Remember, when finding the rate of change of something, weβre actually interested in the derivative. Weβre therefore going to find the derivative of π¦ with respect to π₯ and evaluate it at the point when π₯ is equal to 11π by six. So how do we differentiate this function? Well, we begin by using the definition of the secant function. We know sec π₯ is equal to one over cos π₯. And we can write π¦ as being equal to negative two over cos of two π₯. And then, there are a couple of things we could do. We could rewrite this as negative two times cos two π₯ to the power of negative one and apply the chain rule. Or since this is written as a fraction, we can apply the quotient rule. Remember, this says that if π’ and π£ are differentiable functions, the derivative of that quotient is π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all over π£ squared. The numerator of our fraction is negative two. So we let π’ be equal to negative two. And thus π£ is equal to cos of two π₯.

To use the quotient rule, weβre going to need to find the derivative of each of these. The derivative of a constant is zero. So dπ’ by dπ₯ is equal to zero. And then, we quote a standard result for the derivative of cos of ππ₯. And we see that dπ£ by dπ₯ is equal to negative two times sin two π₯. We can substitute everything we know into the formula for the quotient rule. And we see that dπ¦ by dπ₯ is equal to cos of two π₯ times zero minus negative two times negative two sin of two π₯ all over cos two π₯ squared. This simplifies to negative four sin of two π₯ over cos squared two π₯. Now, we can separate this slightly and write it as negative four sin two π₯ over cos two π₯ times one over cos two π₯. And we recall the identity sin π₯ over cos π₯ equals tan π₯. And we find we have the general result for the derivative of the secant function.

Remember though, we were looking to find the rate of change of π¦ when π₯ is equal to 11π by six. So in fact, weβre going to substitute π₯ is equal to 11π by six into our expression for the derivative. And weβre actually better off leaving it here in terms of sine and cosine. We substitute π₯ is equal to 11π by six. And we get negative four sin of two times 11π by six over cos squared of two times 11π by six, which is eight root three. The rate of change of π¦ when π₯ is equal to 11π by six is eight root three. And this example has demonstrated a general result. We find that the derivative of sec ππ₯ is equal to π sec of ππ₯ times tan of ππ₯. Weβre now going to perform a similar process to help us find the derivative of the cotangent function.

Determine dπ¦ by dπ₯ given that π¦ is equal to negative three cos of four π₯ plus three cot of four π₯.

To find the derivative of π¦ with respect to π₯, we can individually differentiate negative three cos of four π₯ and three cot of four π₯. We know that the derivative of cos of ππ₯ is negative π sin of ππ₯. So we see that the derivative of negative three cos of four π₯ is negative three times negative four sin of four π₯, which is simply 12 sin four π₯. But what about the derivative of three cot of four π₯. Well, cot of π₯ is the same as one over tan π₯. And tan π₯ is equal to sin π₯ over cos of π₯. We can therefore say that cot of four π₯ would be the same as one over tan of four π₯ or one over sin of four π₯ over cos of four π₯. And we obtain that three cot of four π₯ is therefore three over sin of four π₯ over cos of four π₯, which can of course be written as three cos of four π₯ over sin of four π₯.

So really, we need to differentiate three cos of four π₯ over sin of four π₯ with respect to π₯. And weβre going to use the quotient rule to differentiate this. We let π’ be equal to three cos of four π₯. And π£ is equal to sin of four π₯. We know that the derivative of cos of four π₯ is negative four sin of four π₯. So dπ’ by dπ₯ is negative 12 sin of four π₯. We can also use the general result for the derivative of the sine function. And we get dπ£ by dπ₯ to be equal to four cos of four π₯. We can now substitute everything we know about our function into the derivative. And we end up with sin of four π₯ times negative 12 sin of four π₯ minus three cos of four π₯ times four cos of four π₯ over sin squared four π₯. This simplifies to negative 12 sin squared of four π₯ minus 12 cos squared of four π₯ over sin squared four π₯.

But if we factor negative 12, we see that the numerator of this fraction is the same as negative 12 times sin squared four π₯ plus cos squared four π₯. And we can therefore use the identity cos squared π₯ plus sin squared π₯ equals one to find that the derivative of three cot of four π₯ is negative 12 over sin squared of four π₯. Thereβs one more identity we can use. We know that one over sin π₯ is equal to cosec of π₯. And we can write negative 12 over sin squared of four π₯ as negative 12 cosec squared of four π₯. dπ¦ by dπ₯ is the sum of these two results. So itβs equal to 12 sin of four π₯ minus 12 cosec squared of four π₯.

In this example, we demonstrated a result that could be generalized. We saw that the derivative of three cot of four π₯ was negative 12 cosec squared of four π₯. In the same way, we can generalize the result for the derivative of cot ππ₯. Itβs negative π cosec squared of ππ₯. In our next example, weβll look at how to find the derivative of the cosecant function, before considering how our generalised results can help us find the derivatives of more complicated functions.

Given that π¦ is equal to negative 13 cosec of π plus five π₯, find dπ¦ by dπ₯.

To find the derivative of the cosecant function, weβll begin by recalling its definition. We know that cosec of π₯ is equal to one over sin of π₯. This means there are a number of ways that we can find the derivative of our π¦. We could use the trigonometric sum identity. We could differentiate cosec of five π₯ and then consider the transformation that maps cosec of five π₯ onto negative 13 cosec of π plus five π₯. Or alternatively, we could use the chain rule. This says that if π¦ is a function in π’ and π’ itself is a function in π₯, then the derivative of π¦ with respect to π₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯. Weβll let π¦ be equal to negative 13 over sin π’ then, where π’ is equal to π plus five π₯. dπ’ by dπ₯ is simply five. But weβre going to need to use the quotient rule to find dπ¦ by dπ’.

Weβre differentiating π¦ with respect to π’. So Iβm going to redefine the quotient rule using functions π and π in terms of π’. And we see that the derivative of π over π with respect to π’ is equal to π times dπ by dπ’ minus π times dπ by dπ’ over π squared. This means, in our case, we let π be equal to negative 13 and π be equal to sin π’. The derivative of π with respect to π’ is just zero. And we know the derivative of sin π’ with respect to π’ is cos π’. So the derivative of π¦ with respect to π’ is sin π’ times zero minus negative 13 times cos π’ over sin squared π’, which simplifies to 13 cos π’ over sin squared π’. Iβm going to write this as 13 cos π’ over sin π’ times one over sin π’. And this means dπ¦ by dπ’ can be alternatively written as 13 cot π’ cosec π’.

Letβs substitute everything we have into the formula for the chain rule. Thatβs 13 cot π’ cosec π’ times five or 65 cot π’ cosec π’. We replace π’ with π plus five π₯. And we obtain dπ¦ by dπ₯ to be equal to 65 cot of π plus five π₯ times cosec of π plus five π₯. The examples weβve looked at so far give us the following results for the derivatives of the reciprocal trigonometric functions cotangent, secant, and cosecant. Itβs useful to commit these to memory. But also be aware of and ready to apply their derivation where necessary. Weβll now look at how these results can help us to find the derivative of more complicated functions.

Given π¦ is equal to π₯ plus three times nine π₯ plus cosec π₯, find dπ¦ by dπ₯.

Here, we have an expression which is the product of two functions. Weβre therefore going to use the product rule to calculate dπ¦ by dπ₯. This says that the derivative of the product of two differentiable functions π’ and π£ is π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. We therefore let π’ be equal to π₯ plus three and π£ be equal to nine π₯ plus cosec π₯. The derivative of π₯ plus three is simply one. But what about dπ£ by dπ₯? Well, we know that the derivative of nine π₯ is nine. And the derivative of cosec π₯ is negative cosec π₯ cot π₯. So dπ£ by dπ₯ is equal to nine minus cosec π₯ cot π₯. Letβs substitute what we have into the formula for the product rule. We see that dπ¦ by dπ₯ is equal to π₯ plus three times nine minus cosec π₯ cot π₯ plus nine π₯ plus cosec π₯ times one. We distribute our parentheses and then collect like terms. And we see that dπ¦ by dπ₯ is 18π₯ minus π₯ plus three times cosec π₯ cot π₯ plus cosec π₯ plus 27.

If π¦ is equal to negative nine tan eight π₯ sec eight π₯, find dπ¦ by dπ₯.

Here, we have a function which is itself the product of two differentiable functions. So weβre going to use the products rule. This says that the derivative of the product of two differentiable functions π’ and π£ is π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. So weβll let π’ be equal to negative nine tan eight π₯ and π£ be equal to sec eight π₯. We then quote the general result of the derivative of tan ππ₯ is π sec squared ππ₯. And this means the derivative of negative nine tan eight π₯ is negative nine times eight sec squared eight π₯, which is negative 72 sec squared eight π₯.

We also quote the general result for the derivative of sec ππ₯. Itβs π sec ππ₯ times tan ππ₯, which means that dπ£ by dπ₯ is eight sec eight π₯ times tan eight π₯. We can now substitute everything we know into the formula for the products rule. Itβs π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯, which is negative 72 tan squared eight π₯ sec eight π₯ minus 72 sec cubed eight π₯. Weβll consider one more demonstration of the use of the results of the derivatives of reciprocal trigonometric functions.

Given that π¦ is equal to seven cot five π₯ plus three cosec six π₯ to the negative one, find dπ¦ by dπ₯.

In this example, we have a reciprocal function. We could write this as a fraction and apply the quotient rule. Alternatively, we could use the chain rule. Letβs look at how we might use the chain rule. This says that if π¦ is a function in π’ and π’ itself is a function in π₯, then the derivative of π¦ with respect to π₯ is dπ¦ by dπ’ times dπ’ by dπ₯. We will let π’ be equal to seven cot of five π₯ plus three cosec six π₯. This means that π¦ is equal to π’ to the negative one. The derivative of π¦ with respect to π’ is fairly straightforward. Itβs negative one times π’ to the negative two. Then we quote the derivative for cot ππ₯ as being negative π cosec squared ππ₯ and the derivative of cosec ππ₯ as being negative π cosec ππ₯ cot ππ₯. So we see dπ’ by dπ₯ is equal to negative 35 cosec squared five π₯ minus 18 cosec six π₯ cot six π₯.

The derivative of π¦ with respect to π₯ is the product of these two. And we recall that we can write negative π’ to the negative two as one over negative π’ squared. We then divide through by negative one and replace π’ with seven cot five π₯ plus three cosec six π₯. And weβve obtained dπ¦ by dπ₯ to be equal to 35 cosec squared five π₯ plus 18 cosec six π₯ cot six π₯ over seven cot five π₯ plus three cosec six π₯ all squared.

In this video, weβve seen that we can use the quotient rule to find the derivatives of the reciprocal trigonometric functions, cot of π₯, sec of π₯, and cosec of π₯. We saw that the derivative of the cot of ππ₯ plus π is negative π cosec squared ππ₯ plus π. We learned that the derivative of sec of ππ₯ plus π is π sec of ππ₯ plus π times tan of ππ₯ plus π. And the derivative of cosec of ππ₯ plus π is negative π cosec ππ₯ plus π times cot of ππ₯ plus π. And we saw how we can use these standard results in conjunction with the rules of differentiation to find the derivatives of a large number of functions.