Answer :

GinnyAnswer
To find the derivative [tex]\(\frac{dy}{dx}\)[/tex] of the function [tex]\( y = \frac{e^x + e^{-x}}{e^x - e^{-x}} \)[/tex], we will use the quotient rule for differentiation. The quotient rule states that if we have a function [tex]\( y = \frac{u(x)}{v(x)} \)[/tex], then the derivative [tex]\( \frac{dy}{dx} \)[/tex] is given by:

[tex]\[ \frac{dy}{dx} = \frac{v(x) \frac{du}{dx} - u(x) \frac{dv}{dx}}{(v(x))^2} \][/tex]

Here, [tex]\( u(x) = e^x + e^{-x} \)[/tex] and [tex]\( v(x) = e^x - e^{-x} \)[/tex].

1. Differentiate [tex]\(u(x)\)[/tex]:

[tex]\[ u(x) = e^x + e^{-x} \][/tex]
[tex]\[ \frac{du}{dx} = e^x - e^{-x} \][/tex]

2. Differentiate [tex]\(v(x)\)[/tex]:

[tex]\[ v(x) = e^x - e^{-x} \][/tex]
[tex]\[ \frac{dv}{dx} = e^x + e^{-x} \][/tex]

3. Apply the quotient rule:

[tex]\[ \frac{dy}{dx} = \frac{(e^x - e^{-x})(e^x + e^{-x}) - (e^x + e^{-x})(e^x + e^{-x})}{(e^x - e^{-x})^2} \][/tex]

Let's simplify the numerator:

4. Simplify the terms in the numerator:

[tex]\[ \text{Numerator} = (e^x - e^{-x})(e^x + e^{-x}) - (e^x + e^{-x})^2 \][/tex]

Using the distributive property:

[tex]\[ (e^x - e^{-x})(e^x + e^{-x}) = e^{2x} - e^{-2x} \][/tex]

And,

[tex]\[ (e^x + e^{-x})^2 = e^{2x} + 2 + e^{-2x} \][/tex]

Thus, the numerator becomes:

[tex]\[ e^{2x} - e^{-2x} - e^{2x} - 2 - e^{-2x} = - 2 - 2 = - 2 \][/tex]

Therefore,

[tex]\[ \frac{dy}{dx} = \frac{-2}{(e^x - e^{-x})^2} + 1 \][/tex]

The final answer is:

[tex]\[ \frac{dy}{dx} = \frac{(-e^x - e^{-x})(e^x + e^{-x})}{(e^x - e^{-x})^2} + 1 \][/tex]

We can observe that the final expression matches the derived and simplified result:

[tex]\[ \frac{dy}{dx} = \frac{(-e^x - e^{-x})(e^x + e^{-x})}{(e^x - e^{-x})^2} + 1 \][/tex]

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