Solving Sec Homework: Differentiate Ln(cos(5x)) wrt x

[Roodles01]

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!

 

[th4450]

u = cos(5x)
du/dx = -sin(5x)

Something wrong here~

and sec x = 1 / cos x

 

[Roodles01]

Hmmm!
My table of standard derivatives shows;
if function f(x) = cos(ax),
the derivative f'(x) = -a sin(ax)

In my example a = 5, so -5 sin(5x)

Ah! I see the bit I missed out.

 

[Ray Vickson]

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!

What is the *definition* of tan(w)? Surely your textbook tells you that! If not, try Google.

RGV

 

[Roodles01]

Er! Tan (w)= opposite/adjacent

tan(w) = sin(w)/cos(w)
sec(w) = 1/cos(w)

I haven't yet found a relationship between sec (w) & tan(w) which also involves multiplying it by sin(w).
From the WA page I can, of course, see that it has the relationship as tan(w) = sec(w)sin(w). I would like someone to confirm this, please & say whether this is a standard which I haven't yet encountered.

 

[HallsofIvy]

You have it right there- you originally gave, as the derivative, (1/cos(5x))(-sin(5x)). You now recognize that, because the derivative of 5x is 5, it should be (5)(1/cos(5x)(-sin(5x))= -5(sin(5x)/cos(5x))= -5tan(5x).

 

[Roodles01]

Amazingly easy when pointed out.
. . . 'wood for the trees - again!

Completed my original question
f(x)=sec(5x)ln(cos(5x))
after product rule & then chain rule for different bits.
Thank you all.