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### Proof by First Principles

```
Date: 11/16/2001 at 03:11:24
From: Ewart Xia
Subject: How can I prove by first principle that the derivative of
tan(x) is sec^2(x)?

How would I prove this using lemmas 1 (sinx/x=1) and 2?
```

```
Date: 11/16/2001 at 08:35:15
From: Doctor Jubal
Subject: Re: How can I prove by first principle that the derivative of
tan(x) is sec^2(x)?

Hi Ewart,

Thanks for writing to Dr. Math.

To prove a derivative from first principles, you can use the
definition of the derivative:

df    lim    f(x+h) - f(x)
-- =  h->0  ---------------
dx                 h

Since tan(x) = sin(x) / cos(x), we can use the formal definition of
the derivative to prove the derivatives of sin(x) and cos(x), and then
use the quotient rule to find the derivative of tan(x) from them.

Applying the definition of the derivative to sin(x) gives us:

df            lim  sin(x+h) - sin(x)
-- sin(x) =  h->0 -------------------
dx                         h

From trigonometry, sin(x+h) = sin(x)cos(h) + cos(x)sin(h)

df           lim  sin(x)cos(h) + cos(x)sin(h) - sin(x)
-- sin(x) = h->0 --------------------------------------
dx                              h

We can factor out sin(x) and cos(x) to get the two lemmae you gave:

df                  lim  cos(h) - 1             lim  sin(h)
-- sin(x) = sin(x) h->0 ------------  + cos(x) h->0 --------
dx                           h                         h

And filling in the values for the two limits given by the lemmae
proves that Dx[sin(x)] = cos(x). You can use a very similar proof to
show that Dx[cos(x)] = -sin(x), but I will let you try your hand at
that.

Once we have proven the derivatives of sin(x) and cos(x), we can apply
the derivative quotient rule

d f(x)    g(x)f'(x) - f(x)g'(x)
-- ---- = -----------------------
dx g(x)          [ g(x) ]^2

Since tan(x) = sin(x) / cos(x), plug sin(x) = f(x) and cos(x) = g(x)
along with the appropriate derivatives into the quotient rule, and
then use trigonometry to reduce the resulting expression to sec^2(x).

Feel free to write back if you get stuck and need a more thorough
explanation.

- Doctor Jubal, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus
High School Trigonometry

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