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

It's nice to avoid using the Quotient Rule where unnecessary. It saves time and work.

Using the Product Rule:

Thus,

Simplify further:

For the second derivative, we'll use the quotient rule:

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To find the first and second derivatives of ( \frac{{\ln x}}{{x^2}} ), you can use the quotient rule and the chain rule.

First Derivative: [ \frac{{d}}{{dx}}\left(\frac{{\ln x}}{{x^2}}\right) = \frac{{x^2 \cdot \frac{{d}}{{dx}}(\ln x) - \ln x \cdot \frac{{d}}{{dx}}(x^2)}}{{(x^2)^2}} ]

Second Derivative: [ \frac{{d^2}}{{dx^2}}\left(\frac{{\ln x}}{{x^2}}\right) = \frac{{d}}{{dx}}\left(\frac{{x^2 \cdot \frac{{d}}{{dx}}(\ln x) - \ln x \cdot \frac{{d}}{{dx}}(x^2)}}{{(x^2)^2}}\right) ]

Simplify the expressions using the derivative of ( \ln x ) and ( x^2 ): [ \frac{{d}}{{dx}}(\ln x) = \frac{{1}}{{x}} ] [ \frac{{d}}{{dx}}(x^2) = 2x ]

Then, substitute these into the first and second derivative expressions and simplify further.

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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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