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

Given 𝑦 = (𝑥 + 3)(9𝑥 + csc 𝑥), find d𝑦/d𝑥.

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Video Transcript

Given 𝑦 is equal to 𝑥 plus three times nine 𝑥 plus cosec 𝑥, find d𝑦 by d𝑥.

Here, we have an expression which is the product of two functions. We’re therefore going to use the product rule to calculate d𝑦 by d𝑥. This says that the derivative of the product of two differentiable functions 𝑢 and 𝑣 is 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥. We therefore let 𝑢 be equal to 𝑥 plus three and 𝑣 be equal to nine 𝑥 plus cosec 𝑥. The derivative of 𝑥 plus three is simply one. But what about d𝑣 by d𝑥? Well, we know that the derivative of nine 𝑥 is nine. And the derivative of cosec 𝑥 is negative cosec 𝑥 cot 𝑥. So d𝑣 by d𝑥 is equal to nine minus cosec 𝑥 cot 𝑥. Let’s substitute what we have into the formula for the product rule. We see that d𝑦 by d𝑥 is equal to 𝑥 plus three times nine minus cosec 𝑥 cot 𝑥 plus nine 𝑥 plus cosec 𝑥 times one. We distribute our parentheses and then collect like terms. And we see that d𝑦 by d𝑥 is 18𝑥 minus 𝑥 plus three times cosec 𝑥 cot 𝑥 plus cosec 𝑥 plus 27.

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