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

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

y=2x/(x-1)²
y'=[2(x-1)²-2x*2(x-1)]/(x-1)^4
=(2x²-4x+2-4x²+2x)/(x-1)^4
=(-2x²-2x+2)/(x-1)^4
=-2(x²+x-1)/(x-1)^4
y=2/x²+1
k=y'=2*(-2)x^(-3)+1
When x = 1, k = 4 + 1 = 5
Let the tangent equation be y = 5x + M
If the tangent passes (1,1), then M = - 4
So the tangent equation of curve y = 2 / X & # 178; + 1 at point P (1,1) is y = 5x-4

(- 2x to the fourth power) to the fourth power + 2x to the tenth power (- 2x2) 3 + 2x to the tenth power × 5 (x to the fourth power) 3,

Original formula = 16x ^ 16-16x ^ 16 + 10x ^ 14
=10x^14

If 2x + 3 power of 2 - 2x + 1 power of 2 = 96

x=2
First, calculate the constant added to the exponent,
2^3*2^2x-2*2^2x=96
8*2^2x-2*2^2x=96
6*2^2x=96
2^2x=16=2^4
According to the same index, we can get 2x = 4, x = 2