# How do you differentiate #f(x)=ln(e^(x^2+1)/(2x^3-1)^(1/2)) #?

This problem look intimidating but if we can rewrite it using logarithm properties, it actually really easy.

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To differentiate ( f(x) = \ln\left(\frac{e^{x^2+1}}{(2x^3-1)^{1/2}}\right) ), we use the chain rule and the quotient rule. The derivative is:

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

[ f'(x) = \frac{1}{\frac{e^{x^2+1}}{(2x^3-1)^{1/2}}} \cdot \left(\frac{(2x)(e^{x^2+1})(2x^3-1)^{-1/2} - e^{x^2+1}(1/2)(2x^3-1)^{-3/2}(6x^2)}{(2x^3-1)}\right) ]

[ f'(x) = \frac{(2x)(2x^3-1)^{1/2} - 3x(2x^3-1)}{x^2+1} \cdot \frac{1}{(2x^3-1)^{1/2}} ]

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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.

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.

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.

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