Question Video: Differentiating Functions Involving Trigonometric Ratios Using the Quotient Rule | Nagwa Question Video: Differentiating Functions Involving Trigonometric Ratios Using the Quotient Rule | Nagwa

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

## Join Nagwa Classes

If π¦ = (2 sin 3π₯Β²)/π₯Β², find dπ¦/dπ₯.

04:34

### Video Transcript

If π¦ is equal to two times the sin of three π₯ squared all divided by π₯ squared, find the derivative of π¦ with respect to π₯.

In this question, weβre asked to find the derivative of π¦ with respect to π₯. And we can start by noticing that π¦ is the quotient of two differentiable functions. In our numerator, we have the composition of a trigonometric function and a polynomial. And in the denominator, we have a polynomial. So we can differentiate this by using the quotient rule. And we recall the quotient rule tells us if we have two differentiable functions π of π₯ and π of π₯, then we can find the derivative of the quotient π of π₯ over π of π₯ by using the following formula. Itβs equal to π prime of π₯ times π of π₯ minus π prime of π₯ times π of π₯ all divided by π of π₯ all squared. And thatβs provided that π of π₯ is not equal to zero.

Therefore, to use the quotient rule to find the derivative of π¦ with respect to π₯, weβll set π of π₯ to be the function in our numerator β thatβs two times the sin of three π₯ squared β and π of π₯ to be the function in our denominator β thatβs π₯ squared. Then, π¦ is equal to π of π₯ divided by π of π₯. And we can apply the quotient rule.

To apply the quotient rule, we need to find expressions for π prime of π₯ and π prime of π₯. Letβs start with π prime of π₯. Thatβs the derivative of π₯ squared with respect to π₯. We can do this by using the power rule for differentiation. We multiply by the exponent of π₯ and reduce this exponent by one. π prime of π₯ is two π₯. We then need to find an expression for π prime of π₯. However, we can see that π of π₯ is the composition of two functions. Itβs the composition of a trigonometric function and a polynomial. And to differentiate the composition of two functions, we need to recall the chain rule.

The chain rule tells us the derivative of π’ composed with π£ of π₯ is equal to π£ prime of π₯ multiplied by π’ prime evaluated at π£ of π₯. And this is provided that π£ is differentiable at our value of π₯ and π’ is differentiable at π£ of π₯. To apply the chain rule, we need to set our inner function to be π£ of π₯. We can see that the inner function will be three π₯ squared. We can then set our outer function to be π’. π’ evaluated at π£ is two times the sin of π£.

Now, to apply the chain rule, we need to find expressions for π’ prime and π£ prime. Letβs start with π’ prime. We need to find the derivative of two times the sin of π£ with respect to π£. We can recall the derivative of the sine function is the cosine function. So π’ prime of π£ is two cos of π£.

Next, we need to find an expression for π£ prime of π₯. Since π£ of π₯ is a polynomial, weβll do this by using the power rule for differentiation. We multiply by the exponent of π₯ and reduce this exponent by one. π£ prime of π₯ is six π₯. We can then substitute the expressions for π£ prime of π₯, π£ of π₯, and π’ prime into the chain rule to find an expression for the derivative of π of π₯. Substituting our expressions for π’ prime, π£ prime, and π£ of π₯ into the chain rule, we get π prime of π₯ is equal to six π₯ multiplied by two cos of three π₯ squared, which we can simplify to give us 12π₯ cos three π₯ squared.

Now that we found expressions for π prime of π₯ and π prime of π₯, weβre ready to apply the quotient rule, which tells us dπ¦ by dπ₯ is π prime of π₯ times π of π₯ minus π prime of π₯ times π of π₯ all divided by π of π₯ all squared. Substituting in our expressions for π of π₯, π of π₯, π prime of π₯, and π prime of π₯, we get dπ¦ by dπ₯ is equal to 12π₯ cos of three π₯ squared multiplied by π₯ squared minus two π₯ times two sin of three π₯ squared all divided by π₯ squared all squared.

We can simplify this expression term by term. Letβs start with the first term in our numerator. We can group the like terms. π₯ multiplied by π₯ squared is π₯ cubed. So this term simplifies to give us 12π₯ cubed cos three π₯ squared. We can do the same with the second term in our numerator. We can group the like terms. This gives us negative four π₯ sin of three π₯ squared.

Finally, we can use our laws of exponents to simplify our denominator. π₯ squared all squared is π₯ to the power of two times two, which is π₯ to the fourth power. This then gives us the following expression for dπ¦ by dπ₯. And we could leave our answer like this. However, we can simplify this further by dividing both terms in our numerator separately by the denominator. So weβll split the denominator over both terms in our numerator. And we can see in the first term thereβs a shared factor of π₯ cubed in the numerator and denominator. And in the second term, thereβs a shared factor of π₯ in the numerator and denominator. We can then cancel these shared factors to give us our final answer.

The derivative of π¦ is equal to two sin of three π₯ squared all over π₯ squared with respect to π₯ is equal to 12 cos three π₯ squared all over π₯ minus four sin of three π₯ squared all divided by π₯ cubed.

## 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