\(\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-14.93)^2+(-4.93)^2+(-3.31)^2}=16.07\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-14.93}{16.07}=-0.93\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-4.93}{16.07}=-0.31\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-3.31}{16.07}=-0.21\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.93)(4x^{3}y^{4}cos(x^{4})e^{(z^{2})})+(-0.31)(4y^{3}sin(x^{4})e^{(z^{2})})+(-0.21)(2y^{4}zsin(x^{4})e^{(z^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=- 1.24y^{3}sin(x^{4})e^{(z^{2})} - 0.42y^{4}zsin(x^{4})e^{(z^{2})} - 3.72x^{3}y^{4}cos(x^{4})e^{(z^{2})}\end{align*}\)

Find the directional derivative for the function: 

\(\displaystyle f(x,y)=ln(xy)+ye^x\)

in the direction of the vector \(\displaystyle \vec{u}=2\hat{i}+\hat{j}\)

Find the gradient of \(\displaystyle f(x,y)\)

The gradient by definition is the vector:

 \(\displaystyle \vec{\bigtriangledown} f(x,y)=\frac{\partial}{\partial x}f(x,y)\hat{i}+\frac{\partial}{\partial y}f(x,y)\hat{j}\)                             (1)

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Differentiate with respect to \(\displaystyle x\) while holding \(\displaystyle y\) constant: 

\(\displaystyle \frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x}\left[\ln(xy)+ye^x\right ]=\frac{y}{xy}+ye^x\)

\(\displaystyle \frac{\partial}{\partial x}f(x,y)=\frac{1}{x}+ye^x\)

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Differentiate with respect to \(\displaystyle y\) while holding \(\displaystyle x\) constant:

\(\displaystyle \frac{\partial}{\partial y}f(x,y)=\frac{\partial}{\partial y}\left[\ln(xy)+ye^x\right ]=\frac{x}{xy}+e^x\)

\(\displaystyle \frac{\partial}{\partial y}f(x, y)=\frac{1}{y}+e^x\) 

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Enter the components of the gradient, writing in the form of Equation (1) 

\(\displaystyle \vec{\bigtriangledown}f(x,y)=\left(\frac{1}{x}+ye^x\right)\hat{i}+ \left(\frac{1}{y}+e^x\right)\hat{j}\)

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Find the unit vector in the direction of \(\displaystyle \vec{u}\)

\(\displaystyle \vec{u}=2\hat{i}+\hat{j}\)

Divide the vector by its' magnitude \(\displaystyle \left \| \vec{u}\right \|\) to obtain the unit vector \(\displaystyle \hat{u}\), 

\(\displaystyle \hat{u}=\frac{\vec{u}}{\left \| \vec{u}\right \|} =\frac{\vec{u}}{\sqrt{2^2+1^2}}=\frac{2\hat{i}+\hat{j}}{\sqrt{5}}\)

 \(\displaystyle \hat{u}=\frac{\sqrt{5}}{5}(2\hat{i}+\hat{j})\)

You can check the magnitude of the unit vector and see that it is equal to 1 as required. It also has the same direction as the original vector \(\displaystyle \vec{u}\). 

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The directional derivative of \(\displaystyle f(x,y)\) in the direction of the vector \(\displaystyle \hat{u}\) is the dot product of the gradient \(\displaystyle \vec{\bigtriangledown} f(x,y)\) and \(\displaystyle \hat{u}\). Dot products are also referred to as "projections," since the give the component of some vector in the direction of another. 

\(\displaystyle D_uf(x,y)= \vec{\bigtriangledown} f(x,y)\cdot \hat{u}\) 

 \(\displaystyle D_uf(x,y)= \left[ \left(\frac{1}{x}+ye^x\right)\hat{i}+ \left(\frac{1}{y}+e^x\right)\hat{j}\right]\cdot\left[\frac{2\sqrt{5}}{5}\hat{i}+\frac{\sqrt{5}}{5}\hat{j}\right]\)

\(\displaystyle D_uf(x,y) =\frac{\sqrt{5}}{5}\left[\left(\frac{2}{x}+2ye^x\right)+ \left(\frac{1}{y}+e^x\right) \right ]\)

\(\displaystyle D_uf(x,y) =\frac{\sqrt{5}}{5}\left[ e^x(2y+1)+\frac{1}{y}+\frac{2}{x} \right]\)

\(\displaystyle D_uf(x,y) =\frac{\sqrt{5}}{5}\left[ \frac{xye^x(2y+1)+x+2y}{xy}\right]\)