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The positive horizontal coordinate is $x=\sqrt{3/2}=\sqrt{6}/2$. By implicit differentiation,At the selected point this gives $dy/dx=-\sqrt{6}$.

Differentiate the defining function with the other coordinates independent. Put $C=\cosh (x+z)$ for brevity. ThenThe implicit differentiation ratios therefore giveEach expression applies on the implicit surface wherever its own denominator is nonzero. The hyperbolic cosine appears because the derivative of $\sinh (x+z)$ with respect to either $x$ or $z$ is $\cosh (x+z)$.

Implicit differentiation yields $3y^2{y}^{\prime }+2x=0$. At $x=1$, the real curve has $y^3=1$, hence $y=1$. Therefore

By implicit differentiation of the ellipse equation,The tangent line through $(1,1)$ is thereforeIts normal vector is $(9,16)$, parallel to the gradient $(18,32)$ of $9x^2+16y^2$ at that point.

Apply implicit differentiation, remembering that $y$ depends on $x$:Since $1+e^y>0$, division is always legitimate. Therefore

Put $q={\phi }^{\prime }(y-\beta z)$. Apply implicit differentiation with respect to $x$ and $y$, using the chain rule:ConsequentlyAt a point where the assumed differentiable function $z(x,y)$ exists, the first equation forces $\alpha -\beta q\ne 0$. Combining the equations therefore givesThe assertion is understood on such differentiable branches of the implicit function; arbitrary parameter choices need not produce one everywhere.

Regard the refraction angle as a function of the angle of incidence, $\beta =\arcsin (\sin \alpha /n)$. Implicit differentiation of Snell's law gives ${\beta }^{\prime }=\cos \alpha /(n\cos \beta )$. Differentiating the deviation angle gives ${\delta }^{\prime }=2-4{\beta }^{\prime }$. An interior stationary point therefore satisfiesThe stationary deviation of a primary rainbow ray is a minimum: differentiating $\cos \alpha =n\cos \beta {\beta }^{\prime }$ again and using Snell's law gives ${\beta }^{\prime \prime }=\tan \beta (({\beta }^{\prime }{)}^2-1)$. At the stationary point, ${\beta }^{\prime }=1/2$, and hence ${\delta }^{\prime \prime }=-4{\beta }^{\prime \prime }=3\tan \beta >0$ for a positive interior refraction angle. This minimum is responsible for the concentration of emerging light rays near the primary rainbow direction.