How do you do this: d/dx[cos(arctan x)]
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Originally Posted by Amberosia32 How do you do this: d/dx[cos(arctan x)] Have a look at dot point 5 in Krizalid's sig.
cos(arctan(x)) = 1/sqrt(1+x^2) = (1+x^2)^(-1/2) to see this draw a right triangle with an acute angle u whose opposite side is x and whose adjacent side is 1 making the hypteneuse sqrt(1+x^2)
Thank you so much.
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