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kidsasd987
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https://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-b-implicit-differentiation-and-inverse-functions/session-15-implicit-differentiation-and-inverse-functions/MIT18_01SCF10_Ses15b.pdfso derivative of arctan is 1/(x^2+1) and this is obvious since we are considering trigonometry of a triangle of which hypotenuse is sqrt(x^2+1).But can we consider its hypotenuse to be x instead of sqrt(x^2+1)? where cosy=1/x
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