# How do you find the derivative of #y=ln(1/x)#?

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To find the derivative of ( y = \ln(1/x) ), you can use the chain rule and the derivative of the natural logarithm function.

Let's denote ( u = 1/x ). Then, ( y = \ln(u) ).

Now, using the chain rule, the derivative of ( y ) with respect to ( x ) is:

[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} ]

The derivative of ( \ln(u) ) with respect to ( u ) is ( \frac{1}{u} ).

The derivative of ( u = 1/x ) with respect to ( x ) is ( -\frac{1}{x^2} ).

So, substituting these derivatives into the chain rule formula:

[ \frac{dy}{dx} = \frac{1}{u} \cdot (-\frac{1}{x^2}) ]

[ \frac{dy}{dx} = -\frac{1}{xu} ]

[ \frac{dy}{dx} = -\frac{1}{x \cdot \frac{1}{x}} ]

[ \frac{dy}{dx} = -\frac{1}{1} ]

[ \frac{dy}{dx} = -1 ]

Thus, the derivative of ( y = \ln(1/x) ) with respect to ( x ) is ( -1 ).

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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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