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Topi
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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) = (1sin(x)^2)^1/2 = (1y^2)^1/2.
If y = cos(x) then dy/dx = sin(x) and dx/dy = 1/sin(x) = (1cos(2)^2)^1/2 = (1y^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).
 one year ago
 one year ago
Topi Group Title
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) = (1sin(x)^2)^1/2 = (1y^2)^1/2. If y = cos(x) then dy/dx = sin(x) and dx/dy = 1/sin(x) = (1cos(2)^2)^1/2 = (1y^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).
 one year ago
 one year ago

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Evander Group TitleBest ResponseYou've already chosen the best response.0
Look it's happened becouse of this: if "y = cos(x)" and then the derivate will be dy/dx, but as you know the inverse function of y(x) is x(y) here you have to isolate 'x' look it's mean that x = arccos(y) and then the derivatte will be dx/dy= 1/((1x^2))^1/2 that's why....
 one year ago

Topi Group TitleBest ResponseYou've already chosen the best response.0
What I meant that there was no mention of this easy rule that the derivate of invere function is the reciprocal of the derivate of the function: \[\frac{ dx }{ dy }=\frac{ 1 }{ \frac{ dy }{ dx } }\] And so the inverse trigonometric functions are easy to derivate as: If \[y = \sin(x) \rightarrow x = \arcsin(y)\] then \[\frac{ dy }{ dx } = \cos(x) = \sqrt{1  \sin(x)^{2}} =\sqrt{1  y ^{2}}\] and so \[\frac{ dx }{ dy } = \frac{ 1 }{ \sqrt{1  y ^{2}} }\] So just write the derivate dy/dx as a function of y and then take the reciprocal and you have the derivate of the inverse function. This is very easy to do with (hyperbolic) trigonometric functions and exponentiation. Look at this: \[\ y = e^x \rightarrow x = ln(y), \frac{ dy }{ dx } = e^{x} = y \rightarrow \frac{ dx }{ dy } = \frac{ 1 }{ y }\]
 one year ago

Topi Group TitleBest ResponseYou've already chosen the best response.0
Further more as I'm an european we don't use secant very much and so we usually show the derivative of tangent as: \[y = \tan(x), \frac{ dy }{ dx } = \frac{ 1 }{ \cos^2(x) } = 1 + \tan^2(x)\] So to notice that \[1 + \tan^2(x) = 1 + y^2\] we find out that derivate of \[x = \arctan(y)\] is \[\frac{ dx }{ dy } = \frac{ 1 }{ 1 + y^2 }\] No need to draw any triangles to try to find what the derivate of \[x = \arctan(y)\] is.
 one year ago

Evander Group TitleBest ResponseYou've already chosen the best response.0
you're right
 one year ago
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