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How do you find the derivative of the function #y=cos((1-e^(2x))/(1+e^(2x)))#?

Answer 1

#y'=4*e^(2*x)/(1+e^(2+x))^2*sin((1-e^(2x))/(1+e^(2*x)))#

Using the chain rule, then we get #y'=-sin((1-e^(2x))/(1+e^(2x)))*(-2e^(2x)(1+e^(2x))-(1-e^(2x))*e^(2x))/(1+e^(2x))^2#

#y=-sin((1-e^(2x))/(1+e^(2x)))*(2e^(2x)(-1-e^(2x)-1+e^(2x))/(1+e^(2x))^2)# #y'=4e^(2x)/(1+e^(2x))^2*sin((1-e^(2x))/(1+e^(2x)))#

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Answer 2

Christopher Allers

To find the derivative of the function ( y = \cos \left( \frac{1 - e^{2x}}{1 + e^{2x}} \right) ), you would apply the chain rule and the quotient rule.

Here are the steps:

Apply the chain rule: differentiate the outer function (\cos(u)), where (u = \frac{1 - e^{2x}}{1 + e^{2x}}).
Differentiate the inner function (u = \frac{1 - e^{2x}}{1 + e^{2x}}) using the quotient rule.
After finding the derivative of (u), multiply it by the derivative of (\cos(u)) to apply the chain rule.

The final derivative will involve trigonometric functions and exponentials, and it will be complex.

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Answer from HIX Tutor

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.