Determine the equation of the tangent to the curve f(x)=(2x^2 -1)^-5 at x=1

Answer 1

f ' (x) = -5(2x^2 - 1)^(-6) (4x) = -20x/(2x^2 - 1)^6
when x = 1, f ' (1) = -120/1 = -20
f(1) = (2 - 1)^-5 = 1

so you have the point (1,1) and the slope of -120
y-1 = -20(x-1)
y = -20x+21

check:
https://www.wolframalpha.com/input/?i=graph+y+%3D+%282x%5E2+-+1%29%5E-5+%2C+y+%3D+-20x+%2B+21from+0+to+1.5