We're asked to find
the derivative of each function.
Let's begin with f of x.
We are given f of x equals tangent of sine
of negative three x.
Notice here we have a composition of functions.
We actually have a composition of three functions,
but because the outer function is the tangent function,
we will be using the derivative formula.
The derivative of tangent u with respect to x
is equal to secant squared u times u prime.
So u, the inner function, is going to be
sine of negative three x.
Also notice u is also a composite function.
So if we let u equal the sine of negative three x,
we'll also need to find u prime, or du/dx,
and let's find that now.
U prime is equal to the derivative
of sine of negative three x,
and again, we have to apply the chain rule,
where the derivative of sine u with respect to x
is equal to cosine u times u prime.
Notice in our case, now the inner function
is negative three x.
The derivative of sine negative three x with respect to x
is equal to cosine of negative three x
times the derivative of negative three x,
which is negative three.
U prime is equal to negative three cosine negative three x.
So again, we're given f of x equals tangent
of sine of negative three x,
but because we let u equal sine negative three x,
we can think of the function as just tangent u,
and the derivative of tangent u with respect to x
is equal to secant squared u times u prime,
which indicates f prime of x is equal to,
secant squared u is secant squared of sine negative three x,
and then times u prime, where u prime
is negative three cosine negative three x.
So this is the derivative function.
Let's go ahead and write this as f prime of x
equals negative three secant squared
of sine of negative three x
times cosine of negative three x.
I just moved the coefficient of negative three to the front.
And next we're determining the derivative
of g of x equals cosine of cotangent two x.
Notice in this case, the inner function is going to be
cotangent of two x, so we'll let u equal cotangent two x.
To find g prime of x, we'll be using the derivative formula.
The derivative of cosine u with respect to x
is equal to negative sine u times u prime.
Our next step is to determine u prime, or du/dx,
which is the derivative of cotangent two x
with respect to x.
Looking at our formulas, the derivative of cotangent u
with respect to x is equal to
negative cosecant squared u times u prime,
where in our case, the inner function is two x.
The derivative of cotangent two x with respect to x
is equal to negative cosecant squared two x
times the derivative of two x, which is two.
Simplifying, we have negative two cosecant squared two x.
So again, we are given g of x equals
cosine of cotangent of two x.
And again, because we let cotangent two x equal u,
we can think of g of x as just cosine u,
and the derivative of cosine u with respect to x
is negative sine u times u prime,
and therefore g prime of x is equal to negative sine u,
is negative sine of cotangent two x,
and then times u prime, where u prime is
negative two cosecant squared two x.
Let's also re-write this.
Notice how we have a negative times a negative.
Let's write g prime as g prime of x equals
positive two sine of cotangent two x
times cosecant squared of two x.
I hope you found this helpful.