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 • Third Year of Secondary School

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Given that π¦ = 5π₯Β² cot 4π₯, find dπ¦/dπ₯.

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### Video Transcript

Given that π¦ equals five π₯ squared cot four π₯, find dπ¦ by dπ₯.

Looking at our function π¦, we can see that it is a product. Itβs equal to five π₯ squared multiplied by cot four π₯. And so, in order to find its first derivative dπ¦ by dπ₯, weβre going to need to use the product rule of differentiation. This tells us that for two differentiable functions π’ and π£, the derivative with respect to π₯ of their product π’π£ is equal to π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯.

So we can go ahead and find π’ to be one of our factors and π£ to be the other. We then need to find each of their individual derivatives. For the derivative of π’ with respect to π₯, this is relatively straightforward. We can apply the power rule of differentiation, giving five multiplied by two π₯, which is equal to 10π₯. But dπ£ by dπ₯ is a little bit more complicated because our function π£ is a reciprocal trigonometric function. Recall that the cotangent function is the reciprocal of the tangent function. So π£ is equal to one over tan four π₯. And then, we need to consider how weβre going to find this derivative.

Well, one option would be to recall the identity that tan of four π₯ is equal to sin four π₯ over cos four π₯. So the reciprocal of tan four π₯ is cos four π₯ over sin four π₯. And then, we could find this derivative using the quotient rule and standard results for differentiating sine and cosine functions. In fact, though, we donβt need to go through this process every time because we can quote a standard result, which is that the derivative with respect to π₯ of the cotangent of ππ₯ for some constant π is equal to negative π multiplied by csc squared of ππ₯. So we can say that the derivative of cot four π₯ is equal to negative four multiplied by csc squared of four π₯.

This standard result for differentiating the reciprocal trigonometric function cot of ππ₯ is one which we should commit to memory. Although if we canβt recall it, we could rewrite cot of ππ₯ as cos of ππ₯ over sin ππ₯ and derive the result using the quotient rule, as previously discussed. Now, though, weβre ready to substitute into our product rule, we have that dπ¦ by dπ₯ is equal to π’ times dπ£ by dπ₯. Thatβs five π₯ squared multiplied by negative four csc squared four π₯. To this, we then add π£ times dπ’ by dπ₯. Thatβs cot four π₯ multiplied by 10π₯. We can then simplify the coefficient in our first term and reorder the factors in our second term slightly so that weβre not confused into thinking this is cot of four π₯ 10π₯.

Using the product rule then and standard results for differentiating power terms and reciprocal trigonometric functions, we found that for this function π¦, its first derivative dπ¦ by dπ₯ is equal to negative 20π₯ squared csc squared four π₯ plus 10π₯ cot four π₯.

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