The tangent equation of curve y = 1 / X & # 178; + 2 at point (1,3) is

The tangent equation of curve y = 1 / X & # 178; + 2 at point (1,3) is

y=1/x²+2
∴y'=-2/x³
When x = 1, k = f '(1) = - 2
The tangent equation is Y-3 = - 2 (x-1), that is y = - 2x + 5

What is the tangent equation of curve y = 1 / X & # 178; + 2 at point (1,3)?
y=x²-2+2
y'=﹣2x²(-3)
Substituting x = 1
Y '= - 2 × (1 ^ - 3) = - 2, how can this get - 2?

Find the slope of the tangent line at this point first
y'=-2/x³ k=-2
The tangent equation Y-3 = - 2 (x-1), i.e. y = - 2x + 5, is obtained from the point oblique formula

If x and y satisfy the equations 2x + y = 5x + 2Y = 4, then the value of X-Y is ()
A. -1B. 1C. -2D. 2

Method 1: solve the system of equations 2x + y = 5x + 2Y = 4, and get x = 2, y = 1, X-Y = 1. Method 2: subtract the two equations, and get X-Y = 1