Video: Differentiating Functions Involving Trigonometric Ratios Using the Product Rule

Given that 𝑦 = 5π‘₯Β² cot 4π‘₯, find d𝑦/dπ‘₯.

03:12

Video Transcript

Given that 𝑦 equals five π‘₯ squared cot four π‘₯, find d𝑦 by dπ‘₯.

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

So we can go ahead and find 𝑒 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 𝑒 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 d𝑣 by dπ‘₯ 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 π‘₯. 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 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 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 π‘₯ of the cotangent of π‘Žπ‘₯ for some constant π‘Ž is equal to negative π‘Ž multiplied by csc squared of π‘Žπ‘₯. 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‘Žπ‘₯ is one which we should commit to memory. Although if we can’t recall it, we could rewrite cot of π‘Žπ‘₯ as cos of π‘Žπ‘₯ over sin π‘Žπ‘₯ 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 dπ‘₯ is equal to 𝑒 times d𝑣 by dπ‘₯. That’s five π‘₯ squared multiplied by negative four csc squared four π‘₯. To this, we then add 𝑣 times d𝑒 by dπ‘₯. That’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 we’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 𝑦, its first derivative d𝑦 by dπ‘₯ is equal to negative 20π‘₯ squared csc squared four π‘₯ plus 10π‘₯ cot four π‘₯.

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