Equation of tangent line at a point?

Question

Find the equation of the tangent line to the function:

y = [(1-2x^4)^(1/3)] / (x^2-1)^4

at the point (0,1). [Hint: logarithmic differentiation might help you find dy/dx)

Any ideas?


Thanks!

Answer 1

first take the natural log of both sides:
lny = 1/3 ln(1-2x^4) - 4ln(x^2-1)

1/y*y' = 1/2 * 2*4*x^3*(1/(1-2x^4)) - 4*2x*(1/(x^2-1))

y' = y* [1/2 * 2*4*x^3*(1/(1-2x^4)) - 4*2x*(1/(x^2-1))]

at x = 0, y = 1...
y' = 1*[4*0-8*0] = 0
So the slope at (0,1) is 0
So the equation of the line that goes through (0,1) and has a slope of 0 is y = 1

This makes sense if we look at the original formula, which has even powers of "x" only. This means that y(-.0001) = y(.0001), and since the formula is clearly differentiable, then y(0) must be either a minimum or a maximum, and therefore the slope has to be zero.

You don't really need to do the logarithmic differentiation but it does make the algebra involved simpler.

Answer 2

find y'
substitute x=0
slope will be y'
passes through (0,1)
use the slope pointformula for the equation of the tangent