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What is the derivative of #ln[(x(x^2+1)^2)/(2x^3-1)^(1/2)] #?

Answer 1

Isaiah Allen

Use laws of logarithms, letting the function be #f(x)#:

#f(x) = lnx + ln(x^2 + 1)^2 - ln(2x^3 - 1)^(1/2)#

#f(x) = lnx + 2ln(x^2 + 1) - 1/2ln(2x^3 - 1)#

Now use the chain rule on each term.

#f'(x) = 1/x + (4x)/(x^2 + 1) - (6x^2)/(2(2x^3 - 1))#

#f'(x) = 1/x + (4x)/(x^2 + 1) - (3x^2)/(2x^3 - 1)#

Hopefully this helps!

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Answer 2

Christopher Allers

To find the derivative of ( \ln\left[\frac{x(x^2+1)^2}{(2x^3-1)^{1/2}}\right] ), you would apply the chain rule and the rules of logarithmic differentiation. The derivative is:

[ \frac{d}{dx} \left[ \ln\left(\frac{x(x^2+1)^2}{(2x^3-1)^{1/2}}\right) \right] = \frac{1}{\frac{x(x^2+1)^2}{(2x^3-1)^{1/2}}} \cdot \frac{d}{dx} \left(\frac{x(x^2+1)^2}{(2x^3-1)^{1/2}}\right) ]

Now, apply the quotient rule to the expression inside the logarithm. Then, use the chain rule to differentiate the functions within the quotient. Simplify the result if possible.

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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.