Video: Differentiating Trigonometric Functions Using the Chain Rule

If 𝑦 = βˆ’8 sin (sin 6π‘₯) βˆ’ cos (sin 6π‘₯), find d𝑦/dπ‘₯.

03:54

Video Transcript

If 𝑦 equals negative eight sin of sin of six π‘₯ minus cos of sin of six π‘₯, find d𝑦 by dπ‘₯.

We can differentiate this term-by-term. However, each of our terms is a function of a function. And when we want to differentiate a function of a function, we use the chain rule. Let’s remind ourselves of the chain rule. This says that if 𝑦 equals 𝑓 of 𝑒 and 𝑒 equals 𝑔 of π‘₯, then d𝑦 by dπ‘₯ equals d𝑦 by d𝑒 multiplied by d𝑒 by dπ‘₯.

Let’s call this first term 𝑧. And to avoid confusion with the 𝑦 in the question and the 𝑦 in the chain rule formula, let’s replace 𝑦 in the formula with 𝑧. We let our inner function 𝑒 equal sin of six π‘₯. And so, 𝑧 equals negative eight sin of 𝑒. To use the chain rule formula, we need d𝑧 by d𝑒 and d𝑒 by dπ‘₯. Let’s start by finding d𝑧 by d𝑒.

This is going to be the derivative of negative eight sin of 𝑒 with respect to 𝑒. To do this, we remember the main derivatives of trigonometric functions. And we see that the derivative of sin of π‘₯ is cos of π‘₯. And so, the derivative of negative eight sin of 𝑒 with respect to 𝑒 is negative eight cos of 𝑒. And now, we find d𝑒 by dπ‘₯. This is the derivative of sin of six π‘₯ with respect to π‘₯.

If we use the fact that the derivative of sin of π‘Žπ‘₯ equals π‘Ž cos of π‘Žπ‘₯, then d𝑒 by dπ‘₯ equals six cos of six π‘₯. And now, we can apply the formula for the chain rule. Negative eight sin of sin of six π‘₯ differentiates to negative eight cos of 𝑒 multiplied by six cos of six π‘₯. We remember that we let 𝑒 equal sin of six π‘₯. And so, we can replace 𝑒 with sin of six π‘₯.

And now, we move on to our next term. Let’s call this function β„Ž. And so, we replace 𝑧 in our formula for the chain rule with β„Ž. And we’ll replace 𝑒 with 𝑀 because we already assigned 𝑒 in our previous term. So, 𝑀 is our inner, function sin of six π‘₯, which means that β„Ž equals cos of 𝑀. We need to find dβ„Ž by d𝑀 and d𝑀 by dπ‘₯. Let’s start by finding dβ„Ž by d𝑀.

This is the derivative of cos of 𝑀 with respect to 𝑀. We can use the fact that the derivative of cos of π‘₯ with respect to π‘₯ is negative sin of π‘₯. So, dβ„Ž by d𝑀 equals negative sin of 𝑀. We also need to find d𝑀 by dπ‘₯. This is the derivative of sin of six π‘₯ with respect to π‘₯. Using the fact that the derivative of sin of π‘Žπ‘₯ with respect to π‘₯ is π‘Ž cos of π‘Žπ‘₯, d𝑀 by dπ‘₯ equals six cos of six π‘₯. And so, cos of sin of six π‘₯ differentiates to negative sin of 𝑀 multiplied by six cos of six π‘₯.

We remember that we let 𝑀 equal sin of six π‘₯. And so, we replace 𝑀 with sign of six π‘₯. And now, we’ve seen what each term differentiates to, we can put it together to get d𝑦 by dπ‘₯. d𝑦 by dπ‘₯ equals negative eight cos of sin of six π‘₯ multiplied by six cos of six π‘₯ minus, because it was a minus in the question, negative sin of sin of six π‘₯ multiplied by six cos of six π‘₯.

Here, we are subtracting a negative, so we can just write this as an add. And six cos of six π‘₯ appears in both terms, so we can take this out as a common factor. This leaves us with six cos of six π‘₯ all multiplied by negative eight cos of sin of six π‘₯ add sin of sin of six π‘₯.

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