# What is the gradient of cos (xy) - 3y^4 + 3 at point ( 1, pi/2) ?

Feb 19, 2018

$\text{The gradient at the desired point is:} \setminus \quad - \frac{1}{2} \left(\setminus \pi \setminus \hat{i} + \left[2 + 3 \setminus {\pi}^{3}\right] \setminus \hat{j}\right) .$

#### Explanation:

$\text{First, recall the definition of the gradient of a scalar function:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \nabla \left(f \left(x , y\right)\right) \setminus = \setminus \frac{\partial f}{\partial x} \setminus \hat{i} + \frac{\partial f}{\partial y} \setminus \hat{j} \setminus \quad .$

$\text{We are given the function:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad f \left(x , y\right) \setminus = \setminus \cos \left(x y\right) - 3 {y}^{4} + 3 \setminus \quad .$

$\text{We will need to compute" \ \ {del f} / {del x}, \ "and" \ {del f} / {del y}, \ "before finishing.}$
$\text{So let's do that beforehand:}$

$\text{1)} \setminus \quad \frac{\partial f}{\partial x} \setminus = \setminus \frac{\partial}{\partial x} \left(\cos \left(x y\right) - 3 {y}^{4} + 3\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus \frac{\partial}{\partial x} \left[\cos \left(x y\right)\right] - \frac{\partial}{\partial x} \left[3 {y}^{4}\right] + \frac{\partial}{\partial x} \left[3\right]$

$= \setminus - \sin \left(x y\right) \frac{\partial}{\partial x} \left[x y\right] - 0 + 0 \setminus q \quad \setminus q \quad \setminus q \quad \text{we regard y as a constant}$

$= \setminus - \sin \left(x y\right) \setminus \cdot y \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \text{again, we regard y as a constant}$

$= \setminus - y \sin \left(x y\right) .$

$\text{So:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \frac{\partial f}{\partial x} \setminus = \setminus - y \sin \left(x y\right) . \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \left(1\right)$

$\text{And for" \ {del f} / {del y} ":}$

$\text{2)} \setminus \quad \frac{\partial f}{\partial y} \setminus = \setminus \frac{\partial}{\partial y} \left(\cos \left(x y\right) - 3 {y}^{4} + 3\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus \frac{\partial}{\partial y} \left[\cos \left(x y\right)\right] - \frac{\partial}{\partial y} \left[3 {y}^{4}\right] + \frac{\partial}{\partial y} \left[3\right]$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus - \sin \left(x y\right) \frac{\partial}{\partial y} \left[x y\right] - 3 \left[4 {y}^{3}\right] + 0$

$= \setminus - \sin \left(x y\right) \setminus \cdot x - 12 {y}^{3} \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \text{we regard x as a constant}$

$= \setminus - x \sin \left(x y\right) - 12 {y}^{3.}$

$\text{So:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \frac{\partial f}{\partial y} \setminus = \setminus - x \sin \left(x y\right) - 12 {y}^{3.} \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \left(2\right)$

$\text{Putting the results in (1) and (2) together, we get" \ nabla( f(x,y) ) ":}$

$\setminus q \quad \setminus q \quad \setminus \quad \nabla \left(f \left(x , y\right)\right) \setminus = \setminus \frac{\partial f}{\partial x} \setminus \hat{i} + \frac{\partial f}{\partial y} \setminus \hat{j}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus \left[- y \sin \left(x y\right)\right] \setminus \hat{i} + \left[- x \sin \left(x y\right) - 12 {y}^{3}\right] \setminus \hat{j}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus - \left(y \sin \left(x y\right) \setminus \hat{i} + \left[x \sin \left(x y\right) + 12 {y}^{3}\right] \setminus \hat{j}\right) \setminus \quad .$

$\text{Thus:}$

$\setminus q \quad \setminus \quad \setminus \nabla \left(f \left(x , y\right)\right) \setminus = \setminus - \left(y \sin \left(x y\right) \setminus \hat{i} + \left[x \sin \left(x y\right) + 12 {y}^{3}\right] \setminus \hat{j}\right) \setminus \quad .$

$\text{We want the value of this at the point" \ \ ( 1, \pi/2 ). \ "So:}$

$\nabla \left(f \left(x , y\right)\right) {|}_{\left(1 , \setminus \frac{\pi}{2}\right)}$

$\setminus q \quad \setminus q \quad \setminus = \setminus - \left(\setminus \frac{\pi}{2} \sin \left(1 \setminus \cdot \setminus \frac{\pi}{2}\right) \setminus \hat{i} + \left[1 \setminus \cdot \sin \left(1 \setminus \cdot \setminus \frac{\pi}{2}\right) + 12 {\left(\setminus \frac{\pi}{2}\right)}^{3}\right] \setminus \hat{j}\right)$

$\setminus q \quad \setminus q \quad \setminus = \setminus - \left(\setminus \frac{\pi}{2} \sin \left(\setminus \frac{\pi}{2}\right) \setminus \hat{i} + \left[\sin \left(\setminus \frac{\pi}{2}\right) + 12 {\left(\setminus \frac{\pi}{2}\right)}^{3}\right] \setminus \hat{j}\right)$

$\setminus q \quad \setminus q \quad \setminus = \setminus - \left(\left[\setminus \frac{\pi}{2} \setminus \cdot 1\right] \setminus \hat{i} + \left[1 + \left(4 \cdot 3\right) \left(\setminus {\pi}^{3} / {2}^{3}\right)\right] \setminus \hat{j}\right)$

$\setminus q \quad \setminus q \quad \setminus = \setminus - \left(\setminus \frac{\pi}{2} \setminus \hat{i} + \left[1 + 3 \left(\setminus {\pi}^{3} / {2}^{1}\right)\right] \setminus \hat{j}\right)$

$\setminus q \quad \setminus q \quad \setminus = \setminus - \left(\setminus \frac{\pi}{2} \setminus \hat{i} + \left[1 + \frac{3 \setminus {\pi}^{3}}{2}\right] \setminus \hat{j}\right)$

$\setminus q \quad \setminus q \quad \setminus = \setminus - \frac{1}{2} \left(\setminus \pi \setminus \hat{i} + \left[2 + 3 \setminus {\pi}^{3}\right] \setminus \hat{j}\right) \setminus \quad .$

$\text{Thus:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \nabla \left(f \left(x , y\right)\right) {|}_{\left(1 , \setminus \frac{\pi}{2}\right)} \setminus = \setminus - \frac{1}{2} \left(\setminus \pi \setminus \hat{i} + \left[2 + 3 \setminus {\pi}^{3}\right] \setminus \hat{j}\right) \setminus \quad .$

$\text{This is our answer.}$