We want to find the derivative of the given function.
Notice how the function is a quotient,
so we'll have to apply the quotient rule
in order to find this derivative.
Remember we cannot simplify across addition or subtraction,
so it would be wrong to try to simplify
the three and the six, or the hyperbolic sine functions.
So we're going to leave it in its current form
and apply the quotient rule.
So looking at our derivative formula,
the numerator will be equal to function f,
and the denominator will be equal to function g.
So start by setting up the quotient rule.
F prime of x
is going to be equal to...
Notice how the denominator, adjust the denominator squared
so our denominator will be six plus
hyperbolic sine x squared,
and then our numerator is going to be g times f prime
minus f times g prime,
which is the denominator
times the derivative of the numerator
minus the numerator
times the derivative of the denominator.
Now we'll find the derivative here, and here,
and then begin to simplify.
The denominator stays the same,
and it's six plus hyperbolic sine x
times the derivative of three minus hyperbolic sine x,
the derivative of three would be zero,
and the derivative of negative hyperbolic sine x
would be equal to negative hyperbolic cosine x.
Notice how because the inter function is just x,
and the derivative of x is equal to one,
we don't have to apply the chain rule
in order to find this derivative,
even though our derivative formula does
include the chain rule.
Then we'll have minus three minus hyperbolic sine x
times the derivative of six plus hyperbolic sine x.
Derivative of six is zero,
the derivative of hyperbolic sine x is equal to
hyperbolic cosine x.
And now for the next step,
we'll distribute here, and distribute here.
So we'll have f prime of x equals,
again the denominator stays the same
so we'll have negative six hyperbolic cosine x
minus hyperbolic cosine x,
hyperbolic sine x.
You need to be careful here because there's
tracking this product, so we'll have minus three
hyperbolic cosine x,
then we'll have minus negative hyperbolic sine x
hyperbolic cosine x, which will be plus
hyperbolic cosine x,
hyperbolic sine x.
Now we're almost there.
Notice that this term
and this term are like terms,
but they're also opposites,
so these two terms simplify to zero,
so we're left with negative nine hyperbolic cosine x
divided by the quantity six, plus hyperbolic sine x squared.
And that'll do it for this example.
I hope this example was helpful.