The CancelInverses command performs cancellation of the functions that are inverse to each other.

The following simplifications are made:

Any combinations of the previous simplifications are made.  Also, arctan of two arguments is simplified if it contains sin(subexpr).

Note: Not all simplifications are valid everywhere. You should be aware of this when calling CancelInverses, if option 'safe' is given, then only valid simplifications are applied.

$\mathrm{with}\left(\mathrm{SolveTools}\right)\:$

$\mathrm{CancelInverses}\left(1+\exp \left(-x^2\right)\right)$

$1+\frac{1}{{\ⅇ}^{x^2}}$

$\mathrm{CancelInverses}\left(\arccos \left(\cos \left(x-y\cdot 2\right)\right)\right)$

$x-2y$

$\mathrm{CancelInverses}\left(\tan \left(\arctan \left(x-y\cdot 2\right)\right)\right)$

$x-2y$

$\mathrm{CancelInverses}\left(\ln \left(\exp \left(x-y\cdot 2\right)\right)\right)$

$x-2y$

$\mathrm{CancelInverses}\left(\exp \left(\ln \left(x-y\cdot 2\right)\right)\right)$

$x-2y$

$\mathrm{CancelInverses}\left(\ln \left(\mathrm{LambertW}\left(\exp \left(a+b^2+c^3\right)\right)\right)\right)$

$c^3+b^2+a-\mathrm{LambertW}\left({\ⅇ}^{c^3+b^2+a}\right)$

$\mathrm{CancelInverses}\left(\exp \left(\mathrm{LambertW}\left(a+b^2+c^3\right)\right)\right)$

$\frac{c^3+b^2+a}{\mathrm{LambertW}\left(c^3+b^2+a\right)}$

$\mathrm{CancelInverses}\left(g\left(1-{\cos \left(x+y+z\right)}^2\right)\right)$

$g\left({\sin \left(x+y+z\right)}^2\right)$

$\mathrm{CancelInverses}\left({\left(-5\right)}^{\frac{1}{2}}\right)$

$\mathrm{I}\sqrt{5}$

$\mathrm{CancelInverses}\left({\left({\left(x\sin \left(x\right)\right)}^{\frac{1}{2}}\right)}^3\right)$

${\left(x\sin \left(x\right)\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\mathrm{CancelInverses}\left(\exp \left(\ln \left(x\right)+\ln \left(y+z\right)+t\right)\right)$

${\ⅇ}^t\left(y+z\right)x$

$\mathrm{CancelInverses}\left(\exp \left(2\ln \left(x\right)+3\ln \left(y+z\right)-4t\right)\right)$

${\ⅇ}^{-4t}{\left(y+z\right)}^3{x}^2$

$\mathrm{CancelInverses}\left(\exp \left(2\ln \left(\sin \left(x\right)\right)\right)\right)$

${\sin \left(x\right)}^2$

$\mathrm{CancelInverses}\left(\ln \left({\left(\exp \left(ck\right)\right)}^z\exp \left(bt\right)\right)\right)$

$zck+bt$

$\mathrm{CancelInverses}\left(\ln \left(z^{\frac{3}{2}}x\right)\right)$

$\ln \left(x\right)+\frac{3\ln \left(z\right)}{2}$

$\mathrm{CancelInverses}\left(\ln \left({\left(z^{\frac{3}{2}}\right)}^{5}x\right)\right)$

$\ln \left(x\right)+\frac{15\ln \left(z\right)}{2}$

$\mathrm{CancelInverses}\left(\ln \left({\left(\exp \left(z\right)\right)}^{\cos \left(x\right)}\right)\right)$

$\cos \left(x\right)z$

$\mathrm{CancelInverses}\left(\ln \left({\left(s^{f\left(x\right)}\right)}^{g\left(z\right)}\right)\right)$

$g\left(z\right)f\left(x\right)\ln \left(s\right)$

$\mathrm{CancelInverses}\left(x^{\frac{x+1}{\ln \left(x\right)}+\frac{x-1}{\ln \left(x\right)}}\right)$

${\ⅇ}^{2x}$

$\mathrm{CancelInverses}\left(\arctan \left(-\sin \left(b\right)\,\cos \left(b\right)\right)\right)$

$-b$

$\mathrm{CancelInverses}\left(\arctan \left(\cos \left(b\right)\,\sin \left(b\right)\right)\right)$

$\frac{\mathrm{\pi }}{2}-b$

$\mathrm{CancelInverses}\left(\ln \left(\exp \left(x\right)\right)+\cos \left(\arccos \left(x\right)\right)\right)$

$2x$

$\mathrm{CancelInverses}\left(\ln \left(\exp \left(x\right)\right)+\cos \left(\arccos \left(x\right)\right)\,'\mathrm{safe}'\right)$

$\ln \left({\ⅇ}^x\right)+x$