We're asked to determine the derivative
of each given function.
First, we have f of x equals cosecant x times cotangent x.
Whenever we have a product of trig functions,
we should check to see if it will simplify.
In this case, it won't,
cosecant x is equal to one divided by sine x
and cotangent x is equal to cosine x divided by sine x.
So go ahead and leave it in this form.
And because we have a product of two functions of x,
we will have to apply the product rule of differentiation
to find f prime of x.
And therefore, f prime of x
is equal to the first function of cosecant x
times the derivative of the second function,
which is the derivative of cotangent x with respect to x.
Then we have plus the second function
of cotangent x times the derivative of the first function,
which is a derivative of cosecant x with respect to x.
And now we need to find the derivative of cotangent x
and the derivative of cosecant x.
F prime of x is equal to cosecant x
times the derivative of cotangent x,
which is equal to negative cosecant squared x.
And then we have plus cotangent x
times the derivative of cosecant x,
which is equal to negative cosecant x, cotangent x.
And now let's simplify,
f prime of x is equal to negative cosecant cubed x
and then we have plus negative cosecant x
times cotangent squared x.
Let's just write that as minus cosecant x
times cotangent squared x.
And if we wanted to, we could factor this
because we do have a common factor of cosecant x.
Let's also show that form.
Let's go ahead and factor out negative cosecant x,
which gives us negative cosecant x
times the quantity cosecant squared x.
And because we're factoring out negative cosecant x,
we have plus cotangent squared x.
Now, let's find g prime,
given g of x equals negative three cosecant x tangent x.
G prime of x is equal to the first function
of three secant x
times the derivative of the second function,
which is a derivative of tangent x with respect to x
plus the second function of tangent x
times the derivative of the first function,
which is a derivative of three secant x with respect to x.
This gives us g prime of x
is equal to three secant x
times the derivative of tangent x
is equal to secant squared x.
Then we have plus tangent x
times the derivative of three secant x,
which is equal to three secant x, tangent x.
Simplifying, we have g prime of x
is equal to three secant cubed x
and then we have plus three secant x, tangent squared x.
And again, let's go ahead and factor this,
let's factor out the greatest common factor
of three secant x.
This gives us g prime of x
equals three secant x times the quantity secant squared x
plus tangent squared x.
I hope you found this helpful.