Q. 50 - Calculus

Question

Find all points where the first-order partial derivatives of the functions in Exercises $43-54$ are continuous. Then use Theorems 12.28 and 12.31 to determine the sets in which the functions are differentiable.

$f(x,y)=\tan (x+y)$

Step-by-Step Solution

Answer

The answer for the equation  $f(x,y)=\tan (x+y)$ is $\mathrm{x}+\mathrm{y}-\mathrm{z}=\mathrm{\pi }$.

Step1: Given.

Given $f(x,y)=\tan (x+y)$ at point $P=\left(x_0,y_0\right)=(0,\pi )$

The equation of line of tangent is

$$\begin{gathered} f_x\left(x_0,y_0\right)\left(x-x_0\right)+f_y\left(x_0,y_0\right)\left(y-y_0\right)=f\left(x_0,y_0\right) \\ {f}_x\left(0,\pi \right)\left(x-0\right)+f_y\left(0,\pi \right)\left(y-\pi \right)=z-f\left(0,\pi \right) \left(1\right) \end{gathered}$$

Step2: Consider.

$f_x\left(x_0,y_0\right)={\left.\frac{d}{dx}f(x,y)\right|}_{\left(x_0,y_0\right)}={\left.\frac{d}{dx}\tan (x+y)\right|}_{\left(x_0,y_0\right)}$

$f_x\left(x_0,y_0\right)={\left.\frac{d}{dx}\left(\tan \right(x+y\left)\right)\frac{d}{dx}(x+y)\right|}_{\left(x_0,y_0\right)}$

$f_x\left(x_0,y_0\right)={\left.{\sec }^2(x+y)·1\right|}_{\left(x_0,y_0\right)}$

$f_x\left(x_0,y_0\right)={\left.{\sec }^2(x+y)\right|}_{\left(x_0,y_0\right)}$

$f_x(0,\pi )={\left.{\sec }^2(x+y)\right|}_{(0,\pi )}$

$f_x(0,\pi )={\sec }^2(0+\pi )$

$f_x(0,\pi )={\sec }^2\left(\pi \right)$

$f_x(0,\pi )=\frac{1}{{\cos }^2\left(\pi \right)}=\frac{1}{\left(\cos \right(\pi ){)}^2}$

(∵cos(π)=-1)$f_x(0,\pi )=\frac{1}{(-1{)}^2}$                         $(\because \cos (\pi )=-1)$

$f_x(0,\pi )=1$                                                                       $\left(2\right)$

Step3: Considering.

$f_y\left(x_0,y_0\right)={\left.\frac{d}{dy}f(x,y)\right|}_{\left(x_0,y_0\right)}={\left.\frac{d}{dy}\tan (x+y)\right|}_{\left(x_0,y_0\right)}$

$f_y\left(x_0,y_0\right)={\left.\frac{d}{dy}\left(\tan \right(x+y\left)\right)\frac{d}{dy}(x+y)\right|}_{\left(x_0,y_0\right)}$

$f_y\left(x_0,y_0\right)={\left.{\sec }^2(x+y)·1\right|}_{\left(x_0,y_0\right)}$

$f_y\left(x_0,y_0\right)={\left.{\sec }^2(x+y)\right|}_{\left(x_0,y_0\right)}$

$f_y(0,\pi )={\left.{\sec }^2(x+y)\right|}_{(0,\pi )}$

$f_y(0,\pi )={\sec }^2(0+\pi )$

$f_y(0,\pi )={\sec }^2\left(\pi \right)$

$f_y(0,\pi )=\frac{1}{{\cos }^2\left(\pi \right)}=\frac{1}{\left(\cos \right(\pi ){)}^2}$

$f_y(0,\pi )=\frac{1}{(-1{)}^2}$             $(\because \cos \left(\mathrm{\pi }\right)=-1)$

$f_y(0,\pi )=1$                                                                        $\left(3\right)$

Step4: Equation 4 .

$f\left(x_0,y_0\right)=f(0,\pi )=\tan (0+\pi )$

$f(0,\pi )=\tan \left(\pi \right)$

$f(0,\pi )=0$                      $(\because \tan (\mathrm{\pi })=0) \left(4\right)$

Step5: Substituting equation ( 2 ) , ( 3 ) and ( 4 ) in equation ( 1 ) , get.

$$\begin{gathered} 1(x-0)+1(y-\mathrm{\pi })=\mathrm{z}-0 \\ \mathrm{x}-0+\mathrm{y}-\mathrm{\pi }=\mathrm{z} \\ \mathrm{x}+\mathrm{y}-\mathrm{z}=\mathrm{\pi } \end{gathered}$$