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.