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 differenciate y=arcsec x^2+1/x^2-1. and y=squaroot of 1+3tan^2 theta

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differenciate y=arcsec x^2+1/x^2-1. and y=squaroot of 1+3tan^2 theta

 Jun 22, 2015

Best Answer 

 #2
avatar+33652 
+13

Alternative first function

I should note that this result applies for x>0.  The sign changes for x<0.

 Jun 23, 2015

3+0 Answers

 #1
avatar+33652 
+13

Differentiation 1;

Differentiation 2.

.

.
 Jun 23, 2015
 #2
avatar+33652 
+13
Best Answer

Alternative first function

I should note that this result applies for x>0.  The sign changes for x<0.

Alan Jun 23, 2015
 #3
avatar+36 
+5

thank you very much alan for aswering my question(im sorry i didnt notice you had another approach on the 1st problem and yeah that is the right one) , but i got confused on how you wrote the second problem is y= squaroot of 1+ tan^2 (theta) , is it just equal to your y= squaroot of 1+ tan (theta) ^2?

 Jun 23, 2015

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