How do I find #d/dxln (tan (x^2))#?

1 Answer
Gió
Jan 28, 2015

I would use the Chain Rule.
In a way it is like those russian dolls called Matrioskas where you have a smaller doll inside another. To get to the last one you have to open the big ones first.

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(Picture source: https://ghaiaesoterico.wordpress.com/2013/08/20/lenda-de-matrioska/)

Here the smallest doll is the function #x^2# but to get to it you have first to "open" #tan# and #ln#.

To open in this case means to derive.

So you get:

#d/dxln(tan(x^2))=1/(tan(x^2))*1/(cos^2(x^2))*2x#

Remember that:
#d/dxln(x)=1/x#
#d/dxtan(x)=1/(cos^2(x))#
#d/dxx^2=2x#

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