• anonymous
I'm wondering why the derivation of inverse function was not teached as dx/dy = (dy/dx)^-1? This is easy way to calculate the derivate of inverse trigfunctions. If y = sin(x) then dy/dx = cos(x) and thus dx/dy = 1/cos(x) = (1-sin(x)^2)^-1/2 = (1-y^2)^-1/2. If y = cos(x) then dy/dx = -sin(x) and dx/dy = -1/sin(x) = -(1-cos(2)^2)^-1/2 = -(1-y^2)^-1/2. If y = tan(x) then dy/dx = cos(x)^-2 = (cos(x)^2 + sin(x)^2)/cos(x)^2 = 1 + tan(x)^2 and dx/dy = 1/(1 + tan(x)^2) = 1/(1+y^2).
OCW Scholar - Single Variable Calculus
• Stacey Warren - Expert brainly.com
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