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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) = (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).

  • one year ago
  • one year ago

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  1. Evander Group Title
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    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/((1-x^2))^1/2 that's why....

    • one year ago
  2. Topi Group Title
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    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
  3. Topi Group Title
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    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
  4. Evander Group Title
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    you're right

    • one year ago
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