曲线y=1/(x^2)在点(2,1/4)处的切线方程怎么求?

Question

曲线y=1/(x^2)在点(2,1/4)处的切线方程怎么求?

Answer 1

y=x^(-2)
y'=-2x^(-3)
y'(2)=-1/4
x=2,y=1/4
设切线方程为
y=-1/4 x+b
b=1/4+2/4=3/4
则 y=-x/4+3/4