Welcome to a proof
that the derivative of
tangent x with respect to x
is equal to secant squared x.
To begin our proof,
we'll rewrite tangent x
using the quotient identity that tangent x
is equal to sine x divided by cosine x.
And now we'll find the
derivative in this form
by applying the Quotient Rule
which is given below for a reference.
So the derivative of
sine x divide by cosine x
with respect to x is
equal to the denominator,
or cosine x times the
derivative of the numerator,
which would be the derivative
of sine x minus the numerator,
or minus sine x times the
derivative of the denominator,
which is the derivative of cosine x
all divided by the denominator squared,
or in this case cosine x squared.
Now for the next step,
we'll determine our derivatives.
The derivative of sine
x is equal to cosine x
and the derivative of cosine
x is equal to negative sine x.
And now we'll simplify the numerator
and notice how we also
rewrote cosine x squared
as cosine squared x.
So cosine x times cosine
x is cosine squared x.
Then we have minus sine
x times negative sine x
which becomes plus sine squared x
which brings us to this step.
Now, looking at the numerator,
we have cosine squared
x plus sine squared x,
which is equal to one by
this Pythagorian identity,
so this simplifies to one
over cosine squared x.
And because one over cosine
x is equal to secant x,
we know that one over cosine squared x
equals secant squared x
and we have our proof.
The derivative of tangent
x with respect to x
equals secant squared x.
