# Question Video: Differentiating Functions Involving Trigonometric Ratios Using the Product Rule Mathematics • Higher Education

Given that 𝑦 = 5𝑥² cot 4𝑥, find d𝑦/d𝑥.

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### Video Transcript

Given that 𝑦 equals five 𝑥 squared cot four 𝑥, find d𝑦 by d𝑥.

Looking at our function 𝑦, we can see that it is a product. It’s equal to five 𝑥 squared multiplied by cot four 𝑥. And so, in order to find its first derivative d𝑦 by d𝑥, we’re going to need to use the product rule of differentiation. This tells us that for two differentiable functions 𝑢 and 𝑣, the derivative with respect to 𝑥 of their product 𝑢𝑣 is equal to 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥.

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

Well, one option would be to recall the identity that tan of four 𝑥 is equal to sin four 𝑥 over cos four 𝑥. So the reciprocal of tan four 𝑥 is cos four 𝑥 over sin four 𝑥. And then, we could find this derivative using the quotient rule and standard results for differentiating sine and cosine functions. In fact, though, we don’t need to go through this process every time because we can quote a standard result, which is that the derivative with respect to 𝑥 of the cotangent of 𝑎𝑥 for some constant 𝑎 is equal to negative 𝑎 multiplied by csc squared of 𝑎𝑥. So we can say that the derivative of cot four 𝑥 is equal to negative four multiplied by csc squared of four 𝑥.

This standard result for differentiating the reciprocal trigonometric function cot of 𝑎𝑥 is one which we should commit to memory. Although if we can’t recall it, we could rewrite cot of 𝑎𝑥 as cos of 𝑎𝑥 over sin 𝑎𝑥 and derive the result using the quotient rule, as previously discussed. Now, though, we’re ready to substitute into our product rule, we have that d𝑦 by d𝑥 is equal to 𝑢 times d𝑣 by d𝑥. That’s five 𝑥 squared multiplied by negative four csc squared four 𝑥. To this, we then add 𝑣 times d𝑢 by d𝑥. That’s cot four 𝑥 multiplied by 10𝑥. We can then simplify the coefficient in our first term and reorder the factors in our second term slightly so that we’re not confused into thinking this is cot of four 𝑥 10𝑥.

Using the product rule then and standard results for differentiating power terms and reciprocal trigonometric functions, we found that for this function 𝑦, its first derivative d𝑦 by d𝑥 is equal to negative 20𝑥 squared csc squared four 𝑥 plus 10𝑥 cot four 𝑥.