Q) Let $f(x,y,z)=xy^2z^3+ exp(xyz)$. Find the directional derivative of $f$ at the point $(1,1,1)$ in the direction to the point $(-1,2,3)$.
A) I have been given the following formula, $D_u(f(a,b,c))=u \cdot \nabla(f(a,b,c))$, where $u$ is a unit vector.
I first found $u=(-2i+j+2k)/(\sqrt((-2)^2+1^2+2^2)=(-2i+j+2k)/3$. But then I got confused when calculating $\nabla(f(a,b,c))$.
I got $\nabla(f(a,b,c))=(y^2z^3=exp(yz))i+(xz^3+exp(xz))j+(xy^2+exp(xy))$ so at the point (1,1,1) we have $\nabla(f(a,b,c))=(1+e)i+(1+e)j+(1+e)k$.
Thus giving $D_u(f(a,b,c))=(-2i+j+2k)/3\cdot(1+e)i+(1+e)j+(1+e)k=(1+e)/3$. However, I think I have calculated $\nabla(f(a,b,c))$ incorrectly as I don't know how to deal with the exponential.