The exercise says we have to find an expression of a tangent plane to a function that gives us a good approximation of $\alpha =(0.99e^{0.02})^8$. I defined $f(x,y) = ((1-x)e^{2x})^y$, $0 \leq x < 1$, and then calculated $\frac{df}{dx}(0.01,8)$ and $\frac{df}{dy}(0.01,8)$. Both partial derivatives are multiples of $\alpha$. Since $f(0.01,8)$ is obviously a multiple of $\alpha$, I don't get how the tangent plane, which itself makes reference to $\alpha$, is an efficient way to approximate $\alpha$.
Should I just round up the values of $x$? What am I not understanding about this exercise?
Define $f(x,y,z)=(xe^{y} )^{z}=1$ .Then find the gradient of the function when $x=1$, $y=0$ and $z=8$. The gradient is perpendicular to the level surface and gives you the normal vector for the plane.