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

# Question Video: Differentiating Functions Involving Reciprocal Trigonometric Ratios Using the Product Rule Mathematics • Third Year of Secondary School

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Given π¦ = (π₯ + 3)(9π₯ + csc π₯), find dπ¦/dπ₯.

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### Video Transcript

Given π¦ is equal to π₯ plus three times nine π₯ plus cosec π₯, find dπ¦ by dπ₯.

Here, we have an expression which is the product of two functions. Weβre therefore going to use the product rule to calculate dπ¦ by dπ₯. This says that the derivative of the product of two differentiable functions π’ and π£ is π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. We therefore let π’ be equal to π₯ plus three and π£ be equal to nine π₯ plus cosec π₯. The derivative of π₯ plus three is simply one. But what about dπ£ by dπ₯? Well, we know that the derivative of nine π₯ is nine. And the derivative of cosec π₯ is negative cosec π₯ cot π₯. So dπ£ by dπ₯ is equal to nine minus cosec π₯ cot π₯. Letβs substitute what we have into the formula for the product rule. We see that dπ¦ by dπ₯ is equal to π₯ plus three times nine minus cosec π₯ cot π₯ plus nine π₯ plus cosec π₯ times one. We distribute our parentheses and then collect like terms. And we see that dπ¦ by dπ₯ is 18π₯ minus π₯ plus three times cosec π₯ cot π₯ plus cosec π₯ plus 27.

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