Question Video: Differentiating Trigonometric Functions Using the Chain Rule | Nagwa Question Video: Differentiating Trigonometric Functions Using the Chain Rule | Nagwa

# Question Video: Differentiating Trigonometric Functions Using the Chain Rule Mathematics • Second Year of Secondary School

## Join Nagwa Classes

If π¦ = β8 sin (sin 6π₯) β cos (sin 6π₯), find dπ¦/dπ₯.

03:54

### Video Transcript

If π¦ equals negative eight sin of sin of six π₯ minus cos of sin of six π₯, find dπ¦ 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. Letβs remind ourselves of the chain rule. This says that if π¦ equals π of π’ and π’ equals π of π₯, then dπ¦ by dπ₯ equals dπ¦ by dπ’ multiplied by dπ’ by dπ₯.

Letβs call this first term π§. And to avoid confusion with the π¦ in the question and the π¦ in the chain rule formula, letβs replace π¦ in the formula with π§. We let our inner function π’ equal sin of six π₯. And so, π§ equals negative eight sin of π’. To use the chain rule formula, we need dπ§ by dπ’ and dπ’ by dπ₯. Letβs start by finding dπ§ by dπ’.

This is going to be the derivative of negative eight sin of π’ with respect to π’. 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 π’ with respect to π’ is negative eight cos of π’. And now, we find dπ’ 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π₯ equals π cos of ππ₯, then dπ’ 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 π’ multiplied by six cos of six π₯. We remember that we let π’ equal sin of six π₯. And so, we can replace π’ with sin of six π₯.

And now, we move on to our next term. Letβs call this function β. And so, we replace π§ in our formula for the chain rule with β. And weβll replace π’ with π€ because we already assigned π’ in our previous term. So, π€ is our inner, function sin of six π₯, which means that β equals cos of π€. We need to find dβ by dπ€ and dπ€ by dπ₯. Letβs start by finding dβ by dπ€.

This is the derivative of cos of π€ with respect to π€. We can use the fact that the derivative of cos of π₯ with respect to π₯ is negative sin of π₯. So, dβ by dπ€ equals negative sin of π€. We also need to find dπ€ 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π₯ with respect to π₯ is π cos of ππ₯, dπ€ by dπ₯ equals six cos of six π₯. And so, cos of sin of six π₯ differentiates to negative sin of π€ multiplied by six cos of six π₯.

We remember that we let π€ equal sin of six π₯. And so, we replace π€ with sign of six π₯. And now, weβve seen what each term differentiates to, we can put it together to get dπ¦ by dπ₯. dπ¦ 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₯.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions