Find the equation of the tangent to the curve y=(x^2-5)^3 at the point (2,-1)

问题描述:

Find the equation of the tangent to the curve y=(x^2-5)^3 at the point (2,-1)

求导y'=6x(x²-5)²,
f'(2)=12,
所以y=12(x-2)-1=12x-25