91影视

Given Information,$f(x,y)=\tan 鈦�\left(xy\right)$at position $P=\left(x_0,y_0\right)=\left(1,鈭�\frac{蟺}{4}\right)$

The line of tangent equation is

$f_x\left(x_0,y_0\right)\left(x鈭�x_0\right)+f_y\left(x_0,y_0\right)\left(y鈭�y_0\right)=z鈭�f\left(x_0,y_0\right)$

Equation $1$

$f_x\left(1,鈭�\frac{蟺}{4}\right)(x鈭�1)+f_y\left(1,鈭�\frac{蟺}{4}\right)\left(y+\frac{蟺}{4}\right)=z鈭�f\left(1,鈭�\frac{蟺}{4}\right)$

Then,

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

$f_x\left(x_0,y_0\right)={\left.\frac{d}{dx}(\tan 鈦�(xy\left)\right)\frac{d}{dx}\left(xy\right)\right|}_{\left(x_0,y_0\right)}$

$\begin{array}{r}{f}_x\left(x_0,y_0\right)={\left.{\sec }^2鈦�\left(xy\right)脳y\right|}_{\left(x_0,y_0\right)}\\ f_x\left(x_0,y_0\right)={\left.y脳{\sec }^2鈦�\left(xy\right)\right|}_{\left(x_0,y_0\right)}\end{array}$

$f_x\left(1,鈭�\frac{蟺}{4}\right)=y脳{\sec }^2鈦�\left(xy\right)鈭�\left(1,鈭�\frac{蟺}{4}\right)$

$f_x\left(1,鈭�\frac{蟺}{4}\right)=鈭�\frac{蟺}{4}脳{\sec }^2鈦�\left(1脳\left(鈭�\frac{蟺}{4}\right)\right)$

$f_x\left(1,鈭�\frac{蟺}{4}\right)=鈭�\frac{蟺}{4}脳{\sec }^2鈦�\left(鈭�\frac{蟺}{4}\right)$

$f_x\left(1,鈭�\frac{蟺}{4}\right)=鈭�\frac{蟺}{4}\left(\frac{1}{{\cos }^2鈦�\left(鈭�\frac{蟺}{4}\right)}\right)=鈭�\frac{蟺}{4}\left(\frac{1}{{\left(\cos 鈦�\left(\frac{蟺}{4}\right)\right)}^2}\right)$

$f_x\left(1,鈭�\frac{蟺}{4}\right)=鈭�\frac{蟺}{4}\left(\frac{1}{{\left(\frac{\sqrt{2}}{2}\right)}^2}\right)$

$\left(鈭�\cos 鈦�\left(\frac{蟺}{4}\right)=\frac{\sqrt{2}}{2}\right)$

$f_x\left(1,鈭�\frac{蟺}{4}\right)=鈭�\frac{蟺}{4{\left(\frac{\sqrt{2}}{2}\right)}^2}=鈭�\frac{蟺}{4\frac{1}{2}}$

Equation $2$

$f_x\left(1,鈭�\frac{蟺}{4}\right)=鈭�\frac{蟺}{2}$

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

$f_y\left(x_0,y_0\right)={\left.\frac{d}{dy}(\tan 鈦�(xy\left)\right)\frac{d}{dy}\left(xy\right)\right|}_{\left(x_0鈭�y_0\right)}$

$f_y\left(x_0,y_0\right)={\left.x脳{\sec }^2鈦�\left(xy\right)\right|}_{\left(x_0,y_0\right)}$

$f_y\left(1,鈭�\frac{蟺}{4}\right)=1脳{\sec }^2鈦�\left(1脳\left(鈭�\frac{蟺}{4}\right)\right)$

$f_y\left(1,鈭�\frac{蟺}{4}\right)={\sec }^2鈦�\left(鈭�\frac{蟺}{4}\right)$

$f_y\left(1,鈭�\frac{蟺}{4}\right)=\left(\frac{1}{{\cos }^2鈦�\left(鈭�\frac{蟺}{4}\right)}\right)=\left(\frac{1}{{\left(\cos 鈦�\left(\frac{蟺}{4}\right)\right)}^2}\right)$

$f_y\left(1,鈭�\frac{蟺}{4}\right)=\left(\frac{1}{{\left(\frac{\sqrt{2}}{2}\right)}^2}\right)$

$\left(鈭�\cos 鈦�\left(\frac{蟺}{4}\right)=\frac{\sqrt{2}}{2}\right)$

$f_y\left(1,鈭�\frac{蟺}{4}\right)=\frac{1}{{\left(\frac{\sqrt{2}}{2}\right)}^2}=\frac{1}{\frac{1}{2}}$

Equation $3$

$f_y\left(1,鈭�\frac{蟺}{4}\right)=2$

$f\left(x_0,y_0\right)=f\left(1,鈭�\frac{蟺}{4}\right)=\tan 鈦�\left(1脳\left(鈭�\frac{蟺}{4}\right)\right)$

$f\left(1,鈭�\frac{蟺}{4}\right)=\tan 鈦�\left(鈭�\frac{蟺}{4}\right)=鈭�\tan 鈦�\left(\frac{蟺}{4}\right)$

Equation $4$

$f\left(1,鈭�\frac{蟺}{4}\right)=鈭�1$

$\left(鈭�\tan 鈦�\left(\frac{蟺}{4}\right)=1\right)$

Equations$2,3$and$4$are substituted by equation$1$, we get,

$2$is multiplied by two sides, we get,

$鈭�蟺x+4y鈭�2z=2鈭�蟺鈭�蟺$

$鈭�蟺x+4y鈭�2z=2鈭�2蟺$

Finally, we get the result $蟺x鈭�4y+2z=2蟺鈭�2$.