# Video: Using the Chain Rule to Differentiate a Composition of Trigonometric Functions

Let π(π₯) = cos (sin π₯). What is the value of πβ²(π/3) to three decimal places?

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### Video Transcript

Let π of π₯ equal cos of sin π₯. What is the value of π prime of π by three to three decimal places?

The function whose first derivative weβve been asked to evaluate at a particular point is the composition of two trigonometric functions. We take sin of π₯ and then we take the cosine of this value. In order to find its derivative then, we need to recall how to differentiate a composition of functions.

Well, we do this using the chain rule, which tells us that if π is a function of π’ and π’ is itself a function of π₯, then the derivative of π with respect to π₯ is equal to the derivative of π with respect to π’ multiplied by the derivative of π’ with respect to π₯. dπ by dπ₯ equals dπ by dπ’ times dπ’ by dπ₯.

What weβre going to do then is to find a new variable π’ to be the inner part of our composite function. So weβll let π’ equal sin π₯. Then, our function π, which was cos of sin π₯, will become cos of π’. So we have π as a function of π’ and π’ as a function of π₯.

In order to apply the chain rule, we need to find the individual derivatives, dπ’ by dπ₯ and dπ by dπ’. So we need to recall how to differentiate trigonometric functions. We have standard results for differentiating sin π₯ and cos π₯, each of which can be proven from first principles. The derivative with respect to π₯ of sin π₯ is cos π₯. And the derivative with respect to π₯ of cos π₯ is negative sin π₯.

In fact, thereβs a useful cycle that we can use when finding the derivatives of sin π₯, cos π₯, negative sin π₯, and negative cos π₯. We just travel in a clockwise direction around this cycle to find each derivative. So applying these rules or following the cycle, we have that dπ’ by dπ₯ is equal to cos π₯. And dπ by dπ’ will be equal to negative sin π’. Remember the variable here is π’, not π₯.

Substituting into the chain rule then, we have that dπ by dπ₯ is equal to negative sin π’ multiplied by cos π₯. Now, this is in terms of both variables, π’ and π₯. And as this derivative is with respect to π₯, we want it to be in terms of π₯ only. So the final stage here is to undo our substitution. Weβre going to replace π’ with this definition of sin π₯. So we have that dπ by dπ₯ is equal to negative sin of sin π₯ multiplied by cos π₯, which looks a little strange, but no stranger than the function we began with.

Now remember, weβre asked to evaluate this derivative at a particular π₯-value, when π₯ is equal to π by three. So substituting π by three for π₯ and weβll use the notation π prime here rather than dπ by dπ₯, we have that π prime of π by three is equal to negative sin of sin π by three multiplied by cos of π by three.

Now, we are allowed access to a calculator in this question, and we will need it in a moment. But we donβt actually need it yet because π by three is one of those special angles for which we should know its trigonometric ratios, sine, cosine, and tangent, exactly in terms of surds. Sin of π by three is equal to root three over two. And cos of π by three is equal to one-half. So we can recall these values without needing to use our calculators yet.

We have then that π prime of π by three is equal to negative sin root three over two multiplied by one-half. And at this stage, we will need our calculator because we donβt know what the value of sin root three over two is. So making sure our calculators are in radian mode, we can now evaluate this. It gives negative 0.38087 and then the decimal continues.

Finally, we remember that we were asked to give our answer to three decimal places. So rounding up, we have that π prime of π by three is equal to negative 0.381 to three decimal places. The standard results we used in this question were the chain rule, which tells us how to differentiate a composition of functions, and standard results for differentiating sine and cosine.