Gradient of a function in multi-variate calculus please help Find the gradient of the function at the given point. 
$g(x, y) = 3xe^{y/x}$, at point $(3, 0)$.
How do you compute the gradient of this function. Please help me.
 A: Initially OP wrote $g(x,y) = 3xy^{\frac{y}{x}}$ but realized it was incorrect. I have left the initial work for posterity. The actual answer is given in the edit. As per my comment above, we want to take the logarithm of $g$ as standard calculus identities do not apply to this function. Doing so we have:
$$\begin{eqnarray} \log g &=& \log(3xy^{\frac{y}{x}}) \\ &=& \log(3x) + \frac{y}{x}\log(y). \end{eqnarray} $$
Differentiating this with respect to $x$ gives:
$$\frac{\partial\log g}{\partial x} = \frac{1}{g}\frac{\partial g}{\partial x} = \frac{1}{x} - \frac{y}{x^2}\log(y).$$
Multiplying both sides of this by $g$, we get that
$$\frac{\partial g}{\partial x} = 3xy^{\frac{y}{x}}\left(\frac{1}{x}-\frac{y}{x^2}\log(y)\right). $$
Do you see how to do this for $y$? Knowing these two, do you see how to get the gradient?
Edit: Since the actual definition of $g$ is $g(x,y) = 3x\exp\left(\frac{y}{x}\right)$, I will use standard calculus techniques.
$$\frac{\partial g}{\partial x} = \frac{\partial}{\partial x}\left(3x\exp\left(\frac{y}{x}\right)\right) = 3\exp\left(\frac{y}{x}\right) + 3x\exp\left(\frac{y}{x}\right)\left(-\frac{y}{x^2}\right).$$
Above I used product rule to differentiate the separate functions of $x$ and then used chain rule on the exponential. Knowing this, do you see how to get $\frac{\partial g}{\partial y}$?
A: Take the partial with respect to $x$ using the product and chain rules:
$$
\begin{align}
\frac{\partial}{\partial x}3xe^{y/x}
&=3e^{y/x}+3x\left(-\frac{y}{x^2}\right)e^{y/x}\\
&=3\frac{x-y}{x}e^{y/x}
\end{align}
$$
Take the partial with respect to $y$ using the chain rule
$$
\begin{align}
\frac{\partial}{\partial y}3xe^{y/x}
&=3x\left(\frac1x\right)e^{y/x}\\
&=3e^{y/x}
\end{align}
$$
The gradient of $g(x,y)$ is $\left(\frac{\partial g}{\partial x},\frac{\partial g}{\partial y}\right)$. Therefore,
$$
\nabla3xe^{y/x}=\frac3xe^{y/x}(x-y,x)
$$
Plugging in $(x,y)=(3,0)$ gives $\nabla g(3,0)=(3,3)$.
A: Answer is <3,3>. $\frac{\partial g}{\partial x} = 3e^{y/x}$.
