I have a problem while calculating inverse function of tanh(x).
I know it is y = sinh(x)/cosh(x) and then I should express x, but I am stuck with that.
Will you help me with this? thx a lot
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1Write $\tanh(x)=\frac{e^{2x}+1}{e^{2x}-1}$. Then solve for $e^{2x}$ in $y=\frac{e^{2x}+1}{e^{2x}-1}$, take $\ln$ and divide by $2$. – conditionalMethod Nov 18 '19 at 19:17
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Use that $$\tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^{2x}-1}{e^{2x}+1}=y$$ so $$e^{2x}=\frac{y+1}{1-y}$$ It is $$\frac{e^x+e^{-x}}{e^x-e^{-x}}\times \frac {e^x}{e^x}$$
Dr. Sonnhard Graubner
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thx for reply, how did you do that expression from direct sinh(x)/cosh(x) to that for with e^(2x) ? – naruto25 Nov 18 '19 at 22:49