Results 1 to 3 of 3

Math Help - Derivative of Trig functions/ The secant proof

  1. #1
    Member
    Joined
    Sep 2011
    Posts
    106
    Thanks
    1

    Derivative of Trig functions/ The secant proof

    My book says that if I learn the derivatives of sin, tangent, and secant function then I can find the rest of the trig derivatives. To do this I must replacing each function by its coresponding and put a negative on the right hand side of the new derivative equation.So I just want to make sure I have these three basic proofs down.

    I can do the proofs for SinX using the f'(x) limit formula and the Addition formula for sin(a+b)...

    I find the tanX by rewriting it as \frac{sinX}{cosX} and using the quotient rule...

    But when I try to find the derivative of the sec I ran into a little issue as far as factoring and rewriting

    \frac{d}{dx}[secX] = \frac{d}{dx}[\frac{1}{cosX}]
    Quotient rule
    \frac{cosX * f'(1) - (1) * f'(cosX)}{cos^2x}
    = \frac{cosX * 0 - (1) * (-sinX)}{cos^2x}
    = \frac{sinX}{cos^2X}
    My first instinct was to take out the sinX
    sinX*\frac{1}{cos^2X}
    = sinXsec^2X
    But this is wrong and I am suppose to take out \frac{1}{cosX}
    \frac{1}{cosX}*\frac{sinX}{cosX}
    = secXtanX

    This has me frustrated. Am I supposed to avoid a squared trig value unless its for the derivatives of tan and cot.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,786
    Thanks
    1570

    Re: Derivative of Trig functions/ The secant proof

    Quote Originally Posted by delgeezee View Post
    My book says that if I learn the derivatives of sin, tangent, and secant function then I can find the rest of the trig derivatives. To do this I must replacing each function by its coresponding and put a negative on the right hand side of the new derivative equation.So I just want to make sure I have these three basic proofs down.

    I can do the proofs for SinX using the f'(x) limit formula and the Addition formula for sin(a+b)...

    I find the tanX by rewriting it as \frac{sinX}{cosX} and using the quotient rule...

    But when I try to find the derivative of the sec I ran into a little issue as far as factoring and rewriting

    \frac{d}{dx}[secX] = \frac{d}{dx}[\frac{1}{cosX}]
    Quotient rule
    \frac{cosX * f'(1) - (1) * f'(cosX)}{cos^2x}
    = \frac{cosX * 0 - (1) * (-sinX)}{cos^2x}
    = \frac{sinX}{cos^2X}
    My first instinct was to take out the sinX
    sinX*\frac{1}{cos^2X}
    = sinXsec^2X
    But this is wrong and I am suppose to take out \frac{1}{cosX}
    \frac{1}{cosX}*\frac{sinX}{cosX}
    = secXtanX

    This has me frustrated. Am I supposed to avoid a squared trig value unless its for the derivatives of tan and cot.
    \displaystyle \frac{\sin{x}}{\cos^2{x}}, \sin{x}\sec^2{x} and \displaystyle \sec{x}\tan{x} are all equivalent.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    Re: Derivative of Trig functions/ The secant proof

    there's nothing wrong with what you did. sec(x)tan(x) is the traditional way to express it, but many trigonometric expressions have several equivalent forms (they are sort of sneaky that way).

    from a practical point of view, it's often preferrable to not have higher powers if you don't need them, as it makes continued differentiation more involved (you have to use the chain rule). but i'd just file this under "good to know, i'll try to remember that".
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trig proof involving secant and cosine
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: November 6th 2011, 04:44 AM
  2. Replies: 3
    Last Post: February 17th 2011, 06:12 PM
  3. taking the derivative of trig functions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 10th 2010, 01:06 PM
  4. Derivative with trig functions
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 7th 2008, 08:14 PM
  5. Derivative with trig functions
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 6th 2008, 05:26 PM

Search Tags


/mathhelpforum @mathhelpforum