calc derivatives

[Sophiesmark]

This one has me dumbstruck
Calculate the derivative

(-ln(cosx))'

I think I'm supposed to use the chain rule but I cant figure out how to get started, any help appreciated

 

[racuna]

(-ln(cosx))'

Just think of it as a composition function.

u=cosx
du/dx=-sinx

y=ln(u)
dy/du=1/u

Then apply the chain rule as follows:
y'=du/dx*dy/du

y'=-(-sinx)*1/u

replace cosx for u and simplify to get y'=sinx/cosx, or tanx

 

[stapel]

So the function is as follows:

. . . . .f(x) = ln(cos(x))

...and you need to find the derivative, f'(x)?

Yes, you would need to use the Chain Rule. In practical terms, the Chain Rules means to differentiate from the outside in, working your way down to the center (the variable). You differentiate the outside layer (the log, in this case), leaving its argument alone. Then multiply by the derivative of the next layer (the cosine, in this case), leaving its argument along. And so forth.

Ask Dr. Math has an article on this topic, and there is also a really old newsgroup thread that might be slightly useful.

Eliz.

 

[Sophiesmark]

i got tan x

 

[Sophiesmark]

cool!! u did too, thanks for the help!