### 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. Itβs equal to five π₯ squared
multiplied by cot four π₯. And so, in order to find its first
derivative dπ¦ by dπ₯, weβre 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
weβre 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 donβt 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 ππ₯ for some constant π
is equal to negative π multiplied by csc squared of ππ₯. 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 ππ₯ is one which we
should commit to memory. Although if we canβt recall it, we
could rewrite cot of ππ₯ as cos of ππ₯ over sin ππ₯ and derive the result using
the quotient rule, as previously discussed. Now, though, weβre ready to
substitute into our product rule, we have that dπ¦ by dπ₯ is equal to π’ times dπ£
by dπ₯. Thatβs five π₯ squared multiplied
by negative four csc squared four π₯. To this, we then add π£ times dπ’
by dπ₯. Thatβs 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 weβre 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₯.