Question Video: Differentiating Functions Involving Trigonometric Ratios Using the Product Rule and the Chain Rule | Nagwa Question Video: Differentiating Functions Involving Trigonometric Ratios Using the Product Rule and the Chain Rule | Nagwa

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

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If π¦ = 7π₯ sin (5π₯ + 4), find dπ¦/dπ₯.

03:30

### Video Transcript

If π¦ is equal to seven π₯ multiplied by the sin of five π₯ plus four, 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 see that π¦ is the product of two differentiable functions: seven π₯, which is a polynomial, and the sin of five π₯ plus four, which is the composition of a trigonometric function and a linear function. Therefore, we can find the derivative of π¦ with respect to π₯ by using the product rule. We recall the product rule tells us for two differentiable functions π of π₯ and π of π₯, the derivative of their product is equal to π prime of π₯ times π of π₯ plus π of π₯ multiplied by π prime of π₯. Therefore, if we set the function π of π₯ to be seven π₯ and the function π of π₯ to be the sin of five π₯ plus four, then π¦ is equal to the product of π of π₯ and π of π₯. So we can find its derivative by using the product rule.

To apply the product rule, we need to find expressions for π prime of π₯ and π prime of π₯. Letβs start with π prime of π₯. π prime of π₯ is the derivative of seven π₯ with respect to π₯. We could do this by using the power rule for differentiation. However, the slope of a linear function is just the coefficient of π₯, so π prime of π₯ is just equal to seven. We now want to find the derivative of π of π₯ with respect to π₯. However, this is slightly more complicated. π of π₯ is the composition of two functions. Itβs the sin of five π₯ plus four. And we recall that we can find the derivative of the composition of two differentiable functions by using the chain rule.

We recall the chain rule tells us the derivative of π’ 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 valid provided π£ is differentiable at our value of π₯ and π’ is differentiable at the value of π£ of π₯. To apply the chain rule, we need to start by setting our inner function to be π£ of π₯. Thatβs the linear function five π₯ plus four. Then we need to set up outer function to be the function π’. π’ evaluated at π£ is the sin of π£. Therefore, weβve rewritten π of π₯ to be π’ composed with π£ of π₯; π’ is a function in π£, and π£ in turn is a function in π₯. Therefore, we can differentiate this by using the chain rule. We need to find expressions for π£ prime of π₯ and π’ prime of π₯.

So letβs start with π£ prime of π₯. Thatβs the derivative of the linear function five π₯ plus four. And remember, the slope of a linear function is just the coefficient of π₯. So π£ prime of π₯ is five. Letβs now find an expression for π’ prime of π£. Thatβs the derivative of the sin of π£ with respect to π£. And we know the derivative of the sine function is the cosine function. π’ prime of π£ is cos π£. We can now substitute our expressions for π£ prime of π₯, π£ of π₯, and π’ prime of π₯ into the chain rule. This then gives us that π prime of π₯ is five multiplied by the cos of five π₯ plus four. Now that we found expressions for π prime of π₯ and π prime of π₯, weβre ready to use the product rule to find an expression for dπ¦ by dπ₯, which is equal to π prime of π₯ times π of π₯ plus π of π₯ times π prime of π₯.

Substituting in these expressions, we get dπ¦ by dπ₯ is equal to seven sin of five π₯ plus four plus seven π₯ multiplied by five cos of five π₯ plus four. We can then simplify this expression. We get seven sin of five π₯ plus four plus 35π₯ multiplied by the cos of five π₯ plus four. And we can then reorder these two terms to get our final answer. We were able to show if π¦ is equal to seven π₯ multiplied by the sin of five π₯ plus four, then dπ¦ by dπ₯ is equal to 35π₯ multiplied by the cos of five π₯ plus four plus seven sin of five π₯ plus four.

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