Welcome to a proof that the derivative
of sine x with respect
to x equals cosine x.
We'll be using the definition
of the derivative shown here
below where f prime of x equals the limit
as h approaches zero of
the difference quotient.
So applying the definition
of derivative for f of x
equals sine x we'd have the
limit as h approaches zero
of f of the quantity
x plus h would be sine
of the quantity x plus
h, and then we'd have
minus f of x, or minus
sine x, all divided by h.
Now we want to expand
the quantity x plus h
using the sum identity
for sine shown here below
for reference, where sine
of A plus B is equal to
sine A times cosine B plus
cosine A times sine B,
in our case A equals x and B equals h,
so when we expand the quantity x plus h,
we end up with sine x times
cosine h plus cosine x
times sine h.
Now for the next step,
we're going to change
the order of these three
terms in the numerator.
Notice here we wrote cosine x sine h first
followed by minus sine x and then finally
plus sine x cosine h.
Now notice how these two
terms here in the numerator
do share a common factor of sine
so now we're going to
factor out negative sine x
from these two terms but we
won't factor the first term.
So when we do this, notice
how if we factor out
negative sine x from sine x
we're left with positive one.
If we factor out negative
sine x from positive sine x
cosine h, we're left with
negative or minus cosine h,
and of course we can check
this by distributing.
Notice how negative sine x
times one is negative sine x
and negative sine x times
negative cosine h is
sine x cosine h.
And now for the next step
we'll write the single fraction
as a difference of two
fractions where both fractions
have a denominator of h.
So now we have the limit
as h approaches zero
of cosine x sine h divided by h minus
sine x times the quantity one
minus cosine h divided by h.
Now I'll take this line to the next slide
and we have two special
limits hidden within
these fractions.
To better recognize them,
let's write the factor
of cosine x in front of the fraction
and write the factor of sine x
in front of the second fraction.
From here, because we have
a limit of a difference,
we can write this as a
difference of two limits
and because cosine x and
sine x are not effected by h
we can factor cosine x and
sine h out of the two limits.
So here notice how we
factored out cosine x
and we're left with the
limit as h approaches zero
of sine h divided by h
and then we have minus
we factored out sine x
so we have sine x times
the limit as h approaches
zero of one minus cosine h
divided by h.
At this point we need to recognize
we have two special limits
here shown below for reference.
The limit as h approaches
zero of sine h divided by h
is equal to one and the
limit as h approaches zero
of the quantity one minus
cosine h divided by h
is equal to zero.
So we have cosine x times one minus sine x
times zero which simplifies
nicely to just cosine x.
So now we have our proof.
We've proven the derivative
of sine x with respect to x
equals cosine x.
And before we go, let's
take a look at the graph of
sine x and cosine x on
the same coordinate plane.
Here we have f of x equals
sine x graphed in blue
and f prime of x equals
cosine x graphed here in red.
It's pretty amazing that
the cosine function values
give us the slopes of the tangent lines
to our sine function.
Notice how where the derivative function
or the cosine function is equal to zero,
here and here, the slopes
of the tangent lines
to the sine function of
zero, meaning we have
horizontal tangent lines,
notice here we have
horizontal tangent line as well as here.
Notice how where we have
our horizontal tangent line
we have a high point or a
low point on our function
f of x equals sine x.
We'll be talking more about
this in the near future.
I hope you found this helpful.
