# How do you find the derivative of #u=(1+e^v)(1-e^v)#?

looking at the RHS we notice it is difference of squares when multiplied out, so it will be easier to differentiate if we multiply out first, rather than use the product rule then simplify.

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To find the derivative of ( u = (1 + e^v)(1 - e^v) ) with respect to ( v ), use the product rule.

Apply the product rule:

( \frac{du}{dv} = (1 + e^v)'(1 - e^v) + (1 + e^v)(1 - e^v)' )

Differentiate each term using the chain rule:

( (1 + e^v)' = 0 + e^v )

( (1 - e^v)' = 0 - e^v )

Substitute these derivatives back into the product rule:

( \frac{du}{dv} = e^v(1 - e^v) + (1 + e^v)(-e^v) )

Simplify:

( \frac{du}{dv} = e^v - e^{2v} - e^v - e^{2v} )

( \frac{du}{dv} = -2e^v - 2e^{2v} )

Thus, the derivative of ( u ) with respect to ( v ) is ( -2e^v - 2e^{2v} ).

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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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