# Math Help - Derivative of Trig functions/ The secant proof

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.

2. ## Re: Derivative of Trig functions/ The secant proof

Originally Posted by delgeezee
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.

3. ## 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".