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Roodles01
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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!
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