# How do you find the first and second derivative of #ln (x^8)/ x^2#?

By the Product Rule, then, we get,

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To find the first and second derivatives of ( \ln \left( \frac{x^8}{x^2} \right) ), we'll use the quotient rule for differentiation and the chain rule.

First derivative: [ \frac{d}{dx} \left( \ln \left( \frac{x^8}{x^2} \right) \right) = \frac{1}{\frac{x^8}{x^2}} \cdot \frac{d}{dx} \left( \frac{x^8}{x^2} \right) ] [ = \frac{x^2}{x^8} \cdot \left( \frac{d}{dx} (x^8) \cdot x^2 - x^8 \cdot \frac{d}{dx} (x^2) \right) ] [ = \frac{x^2}{x^8} \cdot \left( 8x^7 \cdot x^2 - x^8 \cdot 2x \right) ] [ = \frac{8x^9 - 2x^{10}}{x^8} ] [ = \frac{8 - 2x}{x} ]

Second derivative: [ \frac{d^2}{dx^2} \left( \ln \left( \frac{x^8}{x^2} \right) \right) = \frac{d}{dx} \left( \frac{8 - 2x}{x} \right) ] [ = \frac{d}{dx} \left( 8x^{-1} - 2x^{-2} \right) ] [ = -8x^{-2} + 4x^{-3} ] [ = -\frac{8}{x^2} + \frac{4}{x^3} ]

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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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