Question Video: Finding the First Derivative of a Trigonometric Function Raised to a Negative Exponent | Nagwa Question Video: Finding the First Derivative of a Trigonometric Function Raised to a Negative Exponent | Nagwa

Question Video: Finding the First Derivative of a Trigonometric Function Raised to a Negative Exponent Mathematics • Third Year of Secondary School

Given that 𝑦 = (7 cot 5𝑥 + 3 csc 6𝑥)⁻¹, find d𝑦/d𝑥.

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Video Transcript

Given that 𝑦 is equal to seven cot five 𝑥 plus three cosec six 𝑥 to the negative one, find d𝑦 by d𝑥.

In this example, we have a reciprocal function. We could write this as a fraction and apply the quotient rule. Alternatively, we could use the chain rule. Let’s look at how we might use the chain rule. This says that if 𝑦 is a function in 𝑢 and 𝑢 itself is a function in 𝑥, then the derivative of 𝑦 with respect to 𝑥 is d𝑦 by d𝑢 times d𝑢 by d𝑥. We will let 𝑢 be equal to seven cot of five 𝑥 plus three cosec six 𝑥. This means that 𝑦 is equal to 𝑢 to the negative one. The derivative of 𝑦 with respect to 𝑢 is fairly straightforward. It’s negative one times 𝑢 to the negative two. Then we quote the derivative for cot 𝑎𝑥 as being negative 𝑎 cosec squared 𝑎𝑥 and the derivative of cosec 𝑎𝑥 as being negative 𝑎 cosec 𝑎𝑥 cot 𝑎𝑥. So we see d𝑢 by d𝑥 is equal to negative 35 cosec squared five 𝑥 minus 18 cosec six 𝑥 cot six 𝑥.

The derivative of 𝑦 with respect to 𝑥 is the product of these two. And we recall that we can write negative 𝑢 to the negative two as one over negative 𝑢 squared. We then divide through by negative one and replace 𝑢 with seven cot five 𝑥 plus three cosec six 𝑥. And we’ve obtained d𝑦 by d𝑥 to be equal to 35 cosec squared five 𝑥 plus 18 cosec six 𝑥 cot six 𝑥 over seven cot five 𝑥 plus three cosec six 𝑥 all squared.

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