I had actually a Calculus 1 test on inverse attributes today and also one of the concerns asked "What is the antiderivative that $ an$?".

You are watching: Derivative of ln(cosx)

I know now the the appropriate answer is $ln(sec x) + C$, however the prize I placed was $ln|-cos x| + C$ and I to be wondering if the answer would additionally work. My logic is the the derivative of my answer would be $dfrac 1 cos x sin x$ which would certainly simplify come $ an x$.

Can someone simply tell me if mine answer to be right, and if it was wrong what would the derivative of my answer in reality be.



The derivative that $f(x)=ln|x|$ is $f"(x)=frac1x$. For this reason the derivative that $ln|-cos x|$ is $fracsin x-cos x$. This is because the derivative the $-cos x$ is $sin x$.

So the derivative of your expression is $- an x$.


$$y=ln|-cos x|=lnsqrt(-cos x)^2=frac12ln(-cos x)^2=frac12ln(cos^2 x)$$$$y"=frac12(frac-2cos x sin xcos^2 x)=- an x$$


We have$$ fracddx log(kcosx) = frac-ksinxkcosx $$by applying the chain rule. Everything the value of $k$, the derivative is $- anx$.


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