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Previous Year Question (PYQs)



The derivative of $f(x,y)=x^2+xy$ at $P_0(1,1)$ in the direction of unit vector $\vec{u}=\left(\frac{1}{\sqrt{2}}\right)\hat{i}+\left(\frac{1}{\sqrt{2}}\right)\hat{j}$ is





Solution

$\nabla f=(2x+y, x)$ At $(1,1)$: $\nabla f=(3,1)$ Directional derivative: $D_{\vec{u}}f=(3,1)\cdot\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$ $=\frac{3+1}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}$


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