5:11 AM
Hi@PrashinJeevaganth
The directional derivative of $f$ at $P$ in the direction of $(\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}}\hat{j})$ is $\nabla f_{P}.((\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}}\hat{j})) = (3\hat{i}-4\hat{j}).(\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}}\hat{j}) = -\frac{1}{\sqrt{2}}$
Also why $e$ has to play a role here? I was thinking how you got an 'e'?