Question Video: Differentiating Functions Involving Trigonometric Ratios Using the Quotient Rule Mathematics • Higher Education

If 𝑦 = (2 sin 3π‘₯Β²)/π‘₯Β², find d𝑦/dπ‘₯.

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Video Transcript

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

In this question, we’re asked to find the derivative of 𝑦 with respect to π‘₯. And we can start by noticing that 𝑦 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 𝑓 of π‘₯ and 𝑔 of π‘₯, then we can find the derivative of the quotient 𝑓 of π‘₯ over 𝑔 of π‘₯ by using the following formula. It’s equal to 𝑓 prime of π‘₯ times 𝑔 of π‘₯ minus 𝑔 prime of π‘₯ times 𝑓 of π‘₯ all divided by 𝑔 of π‘₯ all squared. And that’s provided that 𝑔 of π‘₯ is not equal to zero.

Therefore, to use the quotient rule to find the derivative of 𝑦 with respect to π‘₯, we’ll set 𝑓 of π‘₯ to be the function in our numerator β€” that’s two times the sin of three π‘₯ squared β€” and 𝑔 of π‘₯ to be the function in our denominator β€” that’s π‘₯ squared. Then, 𝑦 is equal to 𝑓 of π‘₯ divided by 𝑔 of π‘₯. And we can apply the quotient rule.

To apply the quotient rule, we need to find expressions for 𝑓 prime of π‘₯ and 𝑔 prime of π‘₯. Let’s start with 𝑔 prime of π‘₯. That’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. 𝑔 prime of π‘₯ is two π‘₯. We then need to find an expression for 𝑓 prime of π‘₯. However, we can see that 𝑓 of π‘₯ is the composition of two functions. It’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 𝑒 composed with 𝑣 of π‘₯ is equal to 𝑣 prime of π‘₯ multiplied by 𝑒 prime evaluated at 𝑣 of π‘₯. And this is provided that 𝑣 is differentiable at our value of π‘₯ and 𝑒 is differentiable at 𝑣 of π‘₯. To apply the chain rule, we need to set our inner function to be 𝑣 of π‘₯. We can see that the inner function will be three π‘₯ squared. We can then set our outer function to be 𝑒. 𝑒 evaluated at 𝑣 is two times the sin of 𝑣.

Now, to apply the chain rule, we need to find expressions for 𝑒 prime and 𝑣 prime. Let’s start with 𝑒 prime. We need to find the derivative of two times the sin of 𝑣 with respect to 𝑣. We can recall the derivative of the sine function is the cosine function. So 𝑒 prime of 𝑣 is two cos of 𝑣.

Next, we need to find an expression for 𝑣 prime of π‘₯. Since 𝑣 of π‘₯ is a polynomial, we’ll do this by using the power rule for differentiation. We multiply by the exponent of π‘₯ and reduce this exponent by one. 𝑣 prime of π‘₯ is six π‘₯. We can then substitute the expressions for 𝑣 prime of π‘₯, 𝑣 of π‘₯, and 𝑒 prime into the chain rule to find an expression for the derivative of 𝑓 of π‘₯. Substituting our expressions for 𝑒 prime, 𝑣 prime, and 𝑣 of π‘₯ into the chain rule, we get 𝑓 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 𝑓 prime of π‘₯ and 𝑔 prime of π‘₯, we’re ready to apply the quotient rule, which tells us d𝑦 by dπ‘₯ is 𝑓 prime of π‘₯ times 𝑔 of π‘₯ minus 𝑔 prime of π‘₯ times 𝑓 of π‘₯ all divided by 𝑔 of π‘₯ all squared. Substituting in our expressions for 𝑓 of π‘₯, 𝑔 of π‘₯, 𝑓 prime of π‘₯, and 𝑔 prime of π‘₯, we get d𝑦 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. Let’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 d𝑦 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 we’ll split the denominator over both terms in our numerator. And we can see in the first term there’s a shared factor of π‘₯ cubed in the numerator and denominator. And in the second term, there’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 𝑦 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.

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