# What is the derivative of #f(x)=(pi/x^5)(1/(e^(1/x)-1))#?

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To find the derivative of ( f(x) = \frac{\pi}{x^5}\left(\frac{1}{e^{1/x}-1}\right) ), we will use the product rule and chain rule.

Let ( u(x) = \frac{\pi}{x^5} ) and ( v(x) = \frac{1}{e^{1/x}-1} ).

Using the product rule: ( f'(x) = u'(x)v(x) + u(x)v'(x) ).

Now, let's find the derivatives of ( u(x) ) and ( v(x) ):

( u'(x) = \frac{d}{dx}\left(\frac{\pi}{x^5}\right) = -\frac{5\pi}{x^6} ).

( v'(x) = \frac{d}{dx}\left(\frac{1}{e^{1/x}-1}\right) ).

To find ( v'(x) ), we'll use the chain rule. Let ( g(x) = e^{1/x} ).

( v'(x) = \frac{d}{dx}\left(\frac{1}{g(x)-1}\right) = \frac{-1}{(g(x)-1)^2} \cdot g'(x) ).

Now, find ( g'(x) ):

( g'(x) = \frac{d}{dx}(e^{1/x}) = e^{1/x} \cdot \frac{d}{dx}(1/x) = -\frac{1}{x^2} \cdot e^{1/x} ).

Substitute ( g(x) ) and ( g'(x) ) back into ( v'(x) ):

( v'(x) = \frac{-1}{(e^{1/x}-1)^2} \cdot \left(-\frac{1}{x^2} \cdot e^{1/x}\right) ).

Now, put all the pieces together:

( f'(x) = \frac{-5\pi}{x^6} \cdot \frac{1}{e^{1/x}-1} + \frac{\pi}{x^5} \cdot \frac{-1}{(e^{1/x}-1)^2} \cdot \left(-\frac{1}{x^2} \cdot e^{1/x}\right) ).

Simplify the expression to get the final derivative of ( f(x) ).

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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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