We're given f of x comma y
equals sine of the quantity x plus y
where x equals s to the fifth, t to the sixth
and y equals five s minus six t.
We're asked to find the
partial derivative of f with respect to s
and the partial derivative of f with respect to t,
and we're asked to write the partial derivatives
only in terms of s and t.
Because f is a function of x and y,
and x and y are functions of s and t,
we'll have to apply the chain rules that are below
in order to find these partial derivatives.
Let's go ahead and find this first partial derivative
on the next slide.
To set this up,
the partial derivative of f with respect to s
is equal to the partial of f with respect to x
times the partial of x with respect to s
plus the partial of f with respect to y
times the partial of y with respect to s.
We'll find these partial derivatives, simplify
and then make sure we have a function
in terms of only s and t.
The partial of f with respect to s is equal to
the partial of f with respect to x,
so we're going to differentiate
the sine of the quantity x plus y
with respect to x treating y as a constant.
The partial derivative would be cosine
of the quantity x plus y
times the derivative of x plus y with respect to x,
which would just be one.
Again, we're treating y as a constant.
Then we have times
partial of x with respect to s.
The derivative of s to the fifth t to the sixth
with respect to s,
we have five s to the fourth times t to the sixth.
Then we have plus the partial of f with respect y.
Derivative of sine of the quantity
x plus y with respect to y would be cosine
of the quantity x plus y
times the derivative of x plus y with respect to y
which would just be one.
Again, we're treating x as a constant.
Then we have times the partial y with respect to s,
treating t as a constant,
so the partial y with respect to s would just be five.
Let's first simplify this.
Let's write this as
five
s to the fourth t to the sixth times cosine
of the quantity x plus y, plus here we have
five times cosine of the quantity x plus y.
We now need to make a substitution
for x and y using our equations here.
The partial of f with respect to s is going to be equal to
five s to the fourth t to the sixth times cosine of
x is
s to the fifth
t to the sixth
plus y so we'll have plus five s minus six t.
Then we have plus five times cosine again of x plus y
which we now know is s to the fifth t to the sixth
plus five s
minus six t.
Notice now we do have our partial derivative
with respect to s only in terms of s and t.
We could leave our partial derivative in this form here.
This is perfectly fine.
But notice how we could also factor these two terms
to have a common factor of five cosine of the quantity
s to the fifth t to the sixth
plus five s minus six t.
We could also express in the form
five
times cosine
of the quantity s to the fifth t to the sixth
plus five s
minus six t
and then times, if we factor this out of the first term,
we'd be left with s to the fourth t to the sixth.
If we factor this out of the second term,
notice how we're going to be left with one.
It's not necessary to factor it
but notice how it does look more simplified
in factored form.
Either of these two forms
would be acceptable.
Now let's find the partial of f with respect to t.
The partial derivative of f with respect to t
is equal to the partial of f with respect to x
times partial of x with respect to t
plus the partial of f with respect to y
times the partial of y with respect to t.
Again, we'll follow the same process.
We'll first find these partial derivatives, simplify,
and then make sure the function is a function of s and t.
For the partial of f with respect to x, we already found
that's the derivative of the sine of the quantity x plus y
with respect to x treating y as a constant,
which would be cosine of the quantity x plus y
times the derivative of x plus y with respect to x
treating y as a constant which would be just one,
times the partial of x with respect to t.
Differentiating here with respect to t,
we'd have six s to the fifth t to the fifth.
Then we have plus the partial of f with respect to y
is the derivative of
sine of the quantity x plus y with respect to y
which would be cosine of the quantity x plus y
again times one.
Then times the partial of y with respect to t.
Notice that that would be negative six.
Differentiating here with respect to t,
treating s as a constant.
Let's simplify.
Partial of f with respect to t is equal to
six s to the fifth
t to the fifth
cosine of the quantity x plus y.
Here, because of the negative six, we'd have minus
six cosine
of the quantity x plus y.
We now need to make substitutions again for x and y
just like we did before.
The partial of f with respect to t is equal to
six s to the fifth t to the fifth
times cosine of x which is
s to the fifth t to the sixth
and then plus y, which is plus five s minus six t,
and we have minus six times cosine,
again x plus y is going to be s to the fifth t to the sixth
plus five s
minus six t.
Again, we could leave our partial derivative
in this form here.
Notice how we do have function of s and t,
but we can also factor this because we have a common factor
of six cosine of this quantity here.
Let's also show it in factored form.
Partial of f with respect to t is equal to,
again factoring out six cosine
of the quantity s to the fifth t to the sixth
plus five s minus six t.
If we factored this out of the first term,
notice how we'd be left with s to the fifth
t to the fifth.
If we factor this out of the second term,
we're factoring everything out
which leaves us with a one, so we have minus one.
We can express the partial derivative of f with respect to t
in either of these two forms.
I hope you found this helpful.