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

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

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If π¦ = π₯β΅ sin 5π₯, determine dπ¦/dπ₯.

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### Video Transcript

If π¦ is equal to π₯ to the power of five sin five π₯, determine dπ¦ by dπ₯.

Here, we have a function which is the product of two differentiable functions. We can, therefore, use the product rule to help us evaluate dπ¦ by dπ₯. This says that, for our function π¦ is equal to π’ times π£, the derivative of π¦ with respect to π₯ is equal to π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. Weβre going to let π’ be equal to π₯ to the power of five and π£ be equal to sin five π₯. We begin by evaluating the derivative of π’ with respect to π₯. Itβs five π₯ to the power of four. Similarly, since the derivative of sin ππ₯ is π cos ππ₯, dπ£ by dπ₯ is equal to five cos five π₯.

Then we substitute into the formula. π’ times dπ£ by dπ₯ is π₯ to the power of five times five cos five π₯. And π£ times dπ’ by dπ₯ is sin five π₯ times five π₯ to the power of four. And so dπ¦ by dπ₯, in this case, is five π₯ to the power of five times cos five π₯ plus five π₯ to the power of four times sin five π₯.

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