Proof by First PrinciplesDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/