# Video: Differentiating Composite Functions Involving Trigonometric Ratios Using the Chain Rule

If π¦ = sin π₯ + 9π₯ , and π₯ = 6ππ§, find dπ¦ / dπ§ at π§ = 4.

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### Video Transcript

If π¦ equals sin π₯ plus nine sec π₯ and π₯ equals six ππ§, find dπ¦ by dπ§ at π§ equals four.

Letβs have a closer look at what weβve 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 weβre 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 thatβs 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.

Letβs begin with dπ₯ by dπ§ as itβs more straightforward. Now, remember, π is just a constant. So so itβs 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 thatβs 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, letβs find dπ¦ by dπ₯. And we see that π¦ is the sum of a trigonometric term and also a reciprocal trigonometric term. So weβre 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, weβd need to go all the way back to first principles. So we can simply quote this result. But we must remember that itβs 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, weβre asked to evaluate this derivative at a particular point. Itβs 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 ππ§. 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. Weβre 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 ππ§, then dπ¦ by dπ§ when π§ equals four is six π.