Find the directional derivative of the function at $$ P $$ in the direction of $$ \mathbf{v} $$. $$ h(x, y)=e^{4 x} \sin (9 y), \quad P\left(\frac{1}{4}, \frac{\pi}{18}\right), \mathbf{v}=-\mathbf{i} $$

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We are given
$$
h(x, y)=e^{4x}\sin(9y),
$$
the point
$$
P\left(\frac{1}{4}, \frac{\pi}{18}\right),
$$
and the direction
$$
\mathbf{v}=-\mathbf{i} = (-1, 0).
$$

**Step 1. Compute the gradient**

The gradient of $$ h(x,y) $$ is
$$
\nabla h(x,y)=\left(\frac{\partial h}{\partial x}, \frac{\partial h}{\partial y}\right).
$$

Calculate the partial derivatives:
- For $$ x $$:
  $$
  \frac{\partial h}{\partial x} = \frac{d}{dx}\left(e^{4x}\sin(9y)\right)=4e^{4x}\sin(9y),
  $$
- For $$ y $$:
  $$
  \frac{\partial h}{\partial y} = \frac{d}{dy}\left(e^{4x}\sin(9y)\right)=9e^{4x}\cos(9y).
  $$

**Step 2. Evaluate the gradient at $$ P $$**

Substitute $$ x=\frac{1}{4} $$ and $$ y=\frac{\pi}{18} $$:
- $$ x=\frac{1}{4} $$ gives $$ e^{4x}=e^{4\cdot\frac{1}{4}}=e^1=e $$.
- For the $$ \sin $$ term:
  $$
  \sin(9y)=\sin\left(9\cdot\frac{\pi}{18}\right)=\sin\left(\frac{\pi}{2}\right)=1.
  $$
- For the $$ \cos $$ term:
  $$
  \cos(9y)=\cos\left(9\cdot\frac{\pi}{18}\right)=\cos\left(\frac{\pi}{2}\right)=0.
  $$

Thus,
$$
\frac{\partial h}{\partial x}\bigg|_{P}=4e\quad \text{and}\quad \frac{\partial h}{\partial y}\bigg|_{P}=9e\cdot 0=0.
$$
So,
$$
\nabla h(P)=\left(4e,\,0\right).
$$

**Step 3. Compute the directional derivative**

The directional derivative of $$ h $$ at $$ P $$ in the direction of a unit vector $$ \mathbf{u} $$ is given by
$$
D_{\mathbf{u}}h(P)=\nabla h(P)\cdot\mathbf{u}.
$$

The given vector is
$$
\mathbf{v} = (-1,0),
$$
which is already a unit vector. Then,
$$
D_{\mathbf{v}}h(P)=\left(4e,\,0\right)\cdot(-1,0) = 4e\cdot(-1)+0\cdot 0 = -4e.
$$

**Final Answer:**
$$
-4e.
$$
The directional derivative of $$ h(x, y) $$ at point $$ P\left(\frac{1}{4}, \frac{\pi}{18}\right) $$ in the direction of $$ \mathbf{v} = -\mathbf{i} $$ is $$ -4e $$.

Find the directional derivative of the function at in the direction of .

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