### Video Transcript

If π¦ equals sin π₯ plus nine sec π₯ and π₯ equals six ππ§, find dπ¦ by dπ§ at π§ equals four.

Letβs have a closer look at what weβve been given here. Firstly, we have π¦, which is a function of π₯. And then, we have π₯, which is itself a function of π§. And weβre asked to find dπ¦ by dπ§ for a given value of π§. In order to answer this question then, we need to apply the chain rule, which tells us that if π¦ is a function of π₯ and π₯ is a function of π§ β so thatβs exactly the setup we have here β then dπ¦ by dπ§ is equal to dπ¦ by dπ₯ multiplied by dπ₯ by dπ§. We therefore need to find expressions for these two derivatives, dπ¦ by dπ₯ and dπ₯ by dπ§. And we can then multiply them together.

Letβs begin with dπ₯ by dπ§ as itβs more straightforward. Now, remember, π is just a constant. So so itβs six π. And using the power rule of differentiation, we then find that the derivative of π₯ with the respect to π§ is simply six π. You may see that more easily if you think of π§ as π§ to the first power. So when we apply the power rule of differentiation β thatβs multiplying by the exponent and reducing the exponent by one β we get six π multiplied by one multiplied by π§ to the power of zero. But π§ to the power of zero is simply one. So this is six π multiplied by one multiplied by one, which is just six π.

Next, letβs find dπ¦ by dπ₯. And we see that π¦ is the sum of a trigonometric term and also a reciprocal trigonometric term. So weβre going to need to recall the standard results for differentiating each of these. Firstly, we recall that the derivative with respect to π₯ of sin π₯ is cos π₯. And in order to see this, weβd need to go all the way back to first principles. So we can simply quote this result. But we must remember that itβs only valid when the angle is measured in radians.

Secondly, we recall that the derivative with respect to π₯ of the reciprocal trigonometric function sec π₯ is sec π₯ tan π₯. Now, we can just quote this standard result. But in order to remind ourselves where it comes from, we can recall that sec π₯ is equal to one over cos π₯. And we can then apply the quotient rule to differentiating this fraction, with our function π’ in the numerator equal to one and our function π£ in the denominator equal to cos of π₯.

So, differentiating term by term then, we find that dπ¦ by dπ₯ is equal to cos π₯ plus nine sec π₯ tan π₯. Then, to find an expression for dπ¦ by dπ§, we apply the chain rule. And we multiply these two derivatives together, giving six π multiplied by cos π₯ plus nine sec π₯ tan π₯. Now, remember, weβre asked to evaluate this derivative at a particular point. Itβs when π§ is equal to four. But our derivative is in terms of π₯. Before we can do this then, we need to know what π₯ is equal to when π§ is equal to four.

This is relatively straightforward though because we have a simple equation relating π₯ and π§. π₯ is equal to six ππ§. So when π§ is equal to four, π₯ is equal to six π multiplied by four, which is 24π. Substituting this value for π₯ then gives that dπ¦ by dπ§, evaluated when π§ equals four or when π₯ equals 24π, is equal to six π multiplied by cos of 24π plus nine sec of 24π tan of 24π.

Now, either using a calculator or recalling the shape of our trigonometric graphs or the general properties of our cosine and tangent functions, we should recall that cos of 24π is equal to one. And tan of 24π is equal to zero. Sec of 24π, which remember is one over cos of 24π, is one over one, which is also equal to one. So we have six π multiplied by one plus nine multiplied by one multiplied by zero. And of course, that second term inside the parentheses just simplifies to zero. Weβre left with six π multiplied by one, which is simply six π.

So by applying the chain rule and by recalling the standard results for differentiating trigonometric and reciprocal trigonometric functions, we found that if π¦ equals sin π₯ plus nine sec π₯ and if π₯ equals six ππ§, then dπ¦ by dπ§ when π§ equals four is six π.