Asked by Rebecca
Use logarithmic differentiation to find dy/dx.
y = sq. root [(x^2 - 1)/(x^2 + 1)]
y = sq. root [(x^2 - 1)/(x^2 + 1)]
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Answered by
MathMate
y = sq. root [(x^2 - 1)/(x^2 + 1)]
take log (to base e, =ln) on both sides
log(y) = (1/2)(log(x²-1)-log(x²+1)
differentiate with respect to x
dy/dx / y = x/(x²-1)-x/(x²+1)
= 2x/((x²-1)(x²+1))
dy/dx = x/(x²-1)-x/(x²+1)
= 2x/((x²-1)(x²+1)) * √(x²-1)/√(x²+1)
Simplify.
take log (to base e, =ln) on both sides
log(y) = (1/2)(log(x²-1)-log(x²+1)
differentiate with respect to x
dy/dx / y = x/(x²-1)-x/(x²+1)
= 2x/((x²-1)(x²+1))
dy/dx = x/(x²-1)-x/(x²+1)
= 2x/((x²-1)(x²+1)) * √(x²-1)/√(x²+1)
Simplify.
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