Differentiation problem needs solving

I assume you need to find the derivative here?


y=ln(ln tan x)


A bit of the chain rule needs to be applied here. Remember that the derivative of lnx is 1/x and the derivative of tanx is sec^2(x).

y'=1/(ln(tanx)) * 1/tanx * sec^2(x)

=sec^2(x)/(tanx(ln(tanx)))
 
(d/dx)ln(x) = 1/x

(d/dx)tan(x) = [sec(x)]^2

That's about all there is to it.
 
Re: Hmmm...

Makayla52 said:
Could someone explain this bit to me...
Click to expand...
Which "bit"? The part where you didn't post the original question? Or the part where you've posted five ads for your casino site? Or something else?

Be a dear and please be specific. Thanks ever. :roll:

Eliz.
 
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