Why Does Differentiating ln(sec^2(x)) Yield an Unexpected Result?

[trajan22]

Homework Statement

find dy/dx for y=ln(sec^(2)x)

Homework Equations

none

The Attempt at a Solution

1.)1/(sec(x)^2)*dy/dx sec(x)^2

2.)tan(x)/sec(x)^2

3.)i put it in terms of sin and cos...
sin(x)/cos(x)*cos(x)^2/1

4.)i canceled the cos(x) in the denominator and came out with sin(x)cos(x)
the answer i am supposed to get is 2tan(x). where did i go wrong?

 

[neutrino]

The derivative of (sec(x))^2 is NOT tan(x).

Edit: Btw, 1/(sec(x)^2)*dy/dx sec(x)^2, should actually be (1/(sec(x)^2))*d(sec(x)^2)/dx

 

[f(x)]

Solving by chain rule, you get-:

d(ln(sec^2(x)))/d(sec^2(x)) * d(sec^2(x))/d(sec x) * d(sec x)/d(x)

Solving finally, you get 2tanx

 

[trajan22]

oh, right sorry about that on the edit.
and what is the derivative of (sec(x))^2, because i know that the derivative of tan(x) is (sec(x))^2 so i assumed it would work the other way around

 

[f(x)]

oh, right sorry about that on the edit.
and what is the derivative of (sec(x))^2, because i know that the derivative of tan(x) is (sec(x))^2 so i assumed it would work the other way around

no,the other way back is by integrating the derivative to get the original function
derivative of (sec x)^2 is 2secx.secxtanx

 

[trajan22]

oh ok i get it now, thanks for your help. easy mistake

 

[Hurkyl]

Notational fix:

1.)1/(sec(x)^2)*dy/dx sec(x)^2

This is wrong. What you meant to say was something like

[tex]\frac{1}{\sec^2 x} \cdot \frac{d}{dx}\left( \sec^2 x \right)[/tex]

or

[tex]\frac{1}{\sec^2 x} \cdot \frac{d(\sec^2 x)}{dx}[/tex]

or

[tex]\frac{1}{\sec^2 x} \cdot \frac{du}{dx}
\quad \quad (u = \sec^2 x)[/tex]

(In particular, you did not mean to put a y in there. You've already defined [itex]y=\ln \sec^2 x[/itex], and that's certainly not what you wanted in this particular expression)