Find the equation of the line normal to the curve ![\displaystyle y=\tanh \left[\sin^{-1} \left( x-\sqrt{2}\right) \right]](http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+y%3D%5Ctanh+%5Cleft%5B%5Csin%5E%7B-1%7D+%5Cleft%28+x-%5Csqrt%7B2%7D%5Cright%29+%5Cright%5D&bg=ffffff&fg=333333&s=0 "\displaystyle y=\tanh \left[\sin^{-1} \left( x-\sqrt{2}\right) \right]")
at the point where 
.
Solution
We first compute the derivative of the function using the Chain Rule:
![\displaystyle \frac{dy}{dx} = \mathrm{sech}^{2}\left[ \sin^{-1}\left( x-\sqrt{2} \right) \right]\cdot \frac{1}{\sqrt{1-\left(x-\sqrt{2}\right)^{2}}}](http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+%5Cfrac%7Bdy%7D%7Bdx%7D+%3D+%5Cmathrm%7Bsech%7D%5E%7B2%7D%5Cleft%5B+%5Csin%5E%7B-1%7D%5Cleft%28+x-%5Csqrt%7B2%7D+%5Cright%29+%5Cright%5D%5Ccdot+%5Cfrac%7B1%7D%7B%5Csqrt%7B1-%5Cleft%28x-%5Csqrt%7B2%7D%5Cright%29%5E%7B2%7D%7D%7D&bg=ffffff&fg=333333&s=0 "\displaystyle \frac{dy}{dx} = \mathrm{sech}^{2}\left[ \sin^{-1}\left( x-\sqrt{2} \right) \right]\cdot \frac{1}{\sqrt{1-\left(x-\sqrt{2}\right)^{2}}}")
At the point on the curve where 
, 
and the slope of the tangent line is
Thus, the slope of the line normal to the curve at the indicated point is -1. The desired equation is
or
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