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How do I find the derivative of #ln*root3((x-1)/(x+1))#?

Answer 1

#d/dx ln root(3)((x-1)/(x+1))= 2/(3x^2-3) #

# y = ln root(3)((x-1)/(x+1)) # # :. y = ln ((x-1)/(x+1))^(1/3) # # :. y = 1/3ln ((x-1)/(x+1)) # # :. y = 1/3{ln(x-1)-ln(x+1)} # (law of logs)

Differentiating (using the chain rule) gives:

# dy/dx = 1/3{1/(x-1)-1/(x+1)} # # :. dy/dx = 1/3{ ( (x-1)-(x+1) )/((x-1)(x+1)) } # # :. dy/dx = 1/3{ (-2)/(1-x^2) } # # :. dy/dx = 2/3 1/(x^2-1) # # :. dy/dx = 2/(3x^2-3) #

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Answer 2

Paisley Johnson

To find the derivative of ( \ln\left(\sqrt[3]{\frac{x-1}{x+1}}\right) ), we can use the chain rule and the power rule. The derivative is ( \frac{1}{3\sqrt[3]{\left(\frac{x-1}{x+1}\right)^2}(x+1)^2} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.