This is a worked class example I I would like to know how to calculate the directional derivative using the gradient vector formula below is the worked example Thank you!!!! $,f(x,y)=x^{2}xy,P_0(1,2),$vec$(u)=\frac{1}{\sqrt[\phantom{2}]{2}}$vec$()\frac{1}{\sqrt[\phantom{2}]{2}}$vec$()x_0=1,y_0=2$

Solution: Apply the definition

$(\frac{df}{ds})vec(u{)}_1{p}_0=\underset{s0}{lim}\frac{f(x_0{s}_{01},y_{0}s,s)-f(x_0,y_0)}{s}$

$=\underset{s0}{lim}\frac{f(1s*\frac{1}{\sqrt[\phantom{2}]{2}},2s*\frac{1}{\sqrt[\phantom{2}]{2}})-f(1,2)}{\underset{s0}{lim}}\frac{(1\frac{s}{\sqrt[\phantom{2}]{2}}{)}^2(1\frac{s}{\sqrt[\phantom{2}]{2}})(2\frac{s}{\sqrt[\phantom{2}]{2}})-(1^{2}1*2)}{s}$

$=\underset{s0}{lim}(1\frac{2s}{\sqrt[\phantom{2}]{2}}\frac{s^2}{2})(2\frac{3s}{\sqrt[\phantom{2}]{2}}\frac{s^2}{2})-3$

$=\underset{s0}{lim}\frac{\frac{5}{\sqrt[\phantom{2}]{2}}s{s}^2}{s}=\underset{s0}{lim}(\frac{5}{\sqrt[\phantom{2}]{2}}s)=\frac{s}{\sqrt[\phantom{2}]{2}}$