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]
