- #1

Roodles01

- 128

- 0

## Homework Statement

Having started to differentiate Ln(cos(5x)) wrt x I checked the answer with WolframAlpha & got a different method & answer too.

## Homework Equations

differentiate

ln(cos(5x))

## The Attempt at a Solution

I used the Chain rule d(ln(cos(5x)))/dx = d(ln(u))/du * du/dx

Where u = cos(5x)

d(ln(u))/du = 1/u which I have in a table of standard derivatives

= 1/(cos(5x)

du/dx = -sin(5x) again, from a table of standard derivatives.

So according to me d(ln(cos(5x)))/dx = 1/cos(5x) * -sin(5x)

WolframAlpha goes a different route & has the answer

**-5 tan(5x)**

I see how this differs from mine so;

1. why am I wrong,

2. how did WA get from sec(5x)(sin(5x)(-(d(5x)/dx))) to -5tan(5x). i.e how is tan defined in terms of sec?

Sorry to be dense!

Last edited: