Differentiate.

y= u(a cos u + b cot u)

I am not sure if this correct. I was not sure whether to leave the variables as are or where the negatives should be placed?

(u)*(- a sin u - b csc^2 u) + (a cos u + b cot u)*(1)

well, ok, if you are taking the diff with respect to u. I am not certain the author had that in mind.

How else would you go about it?

To differentiate the given equation, you can use the product rule. Let's break down the steps:

Step 1: Identify the terms to differentiate:
The given equation has two terms: "u" and "(a cos u + b cot u)".

Step 2: Apply the product rule:
The product rule states that the derivative of the product of two functions is equal to the derivative of the first function times the second function, plus the first function times the derivative of the second function.

Let's differentiate each term separately:

For the term "u":
The derivative of "u" with respect to "u" is simply 1.

For the term "(a cos u + b cot u)":
To differentiate this term, we consider each sub-term separately.

- The derivative of "a cos u" with respect to "u" can be found using the chain rule. The derivative of "cos u" with respect to "u" is "-sin u", and then we multiply by "a" to get "-a sin u".

- The derivative of "b cot u" with respect to "u" can also be found using the chain rule. The derivative of "cot u" with respect to "u" is "-csc^2 u", and then we multiply by "b" to get "-b csc^2 u".

So, the derivative of the term "(a cos u + b cot u)" with respect to "u" is "-a sin u - b csc^2 u".

Step 3: Combine the derivatives:
Now we can put the derivatives of each term together using the product rule.

The derivative of "y" with respect to "u" will be:
(u)*(-a sin u - b csc^2 u) + (a cos u + b cot u)*(1).

So, the differentiation of the given equation is:
y' = (u)*(-a sin u - b csc^2 u) + (a cos u + b cot u).

To differentiate the given function, y = u(a cos u + b cot u), you can use the product rule and the chain rule.

First, let's apply the product rule:

Let f(u) = u and g(u) = (a cos u + b cot u).

Using the product rule (d/dx [f(u) * g(u)]) = f'(u) * g(u) + f(u) * g'(u), we get:

dy/du = f'(u) * g(u) + f(u) * g'(u)

Now we need to find f'(u) and g'(u):

f'(u) = d/dx[u] = 1

To find g'(u), we need to differentiate (a cos u + b cot u) with respect to u. For this, we'll use the chain rule:

Let h(u) = a cos u and k(u) = b cot u.

Using the chain rule, we get:

g'(u) = h'(u) + k'(u)

To find h'(u), we differentiate a cos u with respect to u:

h'(u) = d/dx[a cos u] = -a sin u

And to find k'(u), we differentiate b cot u with respect to u:

k'(u) = d/dx[b cot u] = - b csc^2 u

Now we can substitute these values back into the original equation:

dy/du = f'(u) * g(u) + f(u) * g'(u)
= (1) * (a cos u + b cot u) + (u) * (-a sin u - b csc^2 u)

Simplifying,

dy/du = a cos u + b cot u - a u sin u - b u csc^2 u

Therefore, the differentiation of y = u(a cos u + b cot u) is
dy/du = a cos u + b cot u - a u sin u - b u csc^2 u.