# How can you prove that #d/dx(cothx) = -csch^2x# using the definition #cothx=coshx/sinhx#?

Please see the proof below

We need

Apply the quotient rule

Therefore,

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Using the definition of hyperbolic cotangent ( \coth(x) = \frac{\cosh(x)}{\sinh(x)} ), we can differentiate ( \coth(x) ) with respect to ( x ) using the quotient rule. Then, recall the derivatives of hyperbolic sine and cosine functions:

[ \frac{d}{dx} \cosh(x) = \sinh(x) ] [ \frac{d}{dx} \sinh(x) = \cosh(x) ]

Now, applying the quotient rule:

[ \frac{d}{dx} \left( \frac{\cosh(x)}{\sinh(x)} \right) = \frac{\sinh(x) \cdot \sinh(x) - \cosh(x) \cdot \cosh(x)}{\sinh^2(x)} ]

Using the identity ( \cosh^2(x) - \sinh^2(x) = 1 ):

[ \frac{\sinh^2(x) - \cosh^2(x)}{\sinh^2(x)} = \frac{1}{\sinh^2(x)} ]

Rewriting in terms of hyperbolic functions:

[ \frac{\sinh^2(x) - \cosh^2(x)}{\sinh^2(x)} = \frac{-\cosh^2(x)}{\sinh^2(x)} ]

Finally, since ( \coth(x) = \frac{\cosh(x)}{\sinh(x)} ):

[ \frac{d}{dx} \coth(x) = \frac{-\cosh^2(x)}{\sinh^2(x)} = -\text{csch}^2(x) ]

Therefore, ( \frac{d}{dx} \coth(x) = -\text{csch}^2(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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