曲线y=1/(3倍根号x^2)在点R(8,1/4)的切线方程是?
问题描述：
曲线y=1/(3倍根号x^2)在点R(8,1/4)的切线方程是?
答
y=1/(3倍根号x^2)=x^(-2/3)
y'=-2/3*x^(-5/3)
k=y'=-2/3*8^(-5/3)=-2/3*2^3*(-5/3)=-2/3*2^(-5)=-1/48
所以切线方程是
y-1/4=-1/48(x-8) 两边乘48
48y-12=-x+8
x+48y-20=0