NG: Let \( f \) be a finction defined by \( f(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \). The derivalive of \( f(x) \) with respect FRE: Soit \( f \) une fonction definio par \( f(x)=\frac{e^{x}-a^{-x}}{e^{x}+e^{-x}} \). La derivee de \( f(x) \) par rapport a \( x \) \( \begin{array}{llll}\text { A. } \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}} & \text { B. } \frac{t}{\left(e^{x}+e^{-x}\right)^{2}} & \text { C. } \frac{e^{x}}{\left(d^{x}+e^{-x}\right)^{2}} & \text { D. } \frac{2}{\left(e^{x}+e^{-x}\right)^{2}}\end{array} \)

To find the derivative of the function \( f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \), we can use the quotient rule. The quotient rule states that if you have a function \( g(x) = \frac{u(x)}{v(x)} \), then the derivative \( g'(x) \) is given by: \[ g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \] In our case, we have: - \( u(x) = e^x - e^{-x} \) - \( v(x) = e^x + e^{-x} \) Now, we need to compute the derivatives \( u'(x) \) and \( v'(x) \): 1. \( u'(x) = e^x + e^{-x} \) 2. \( v'(x) = e^x - e^{-x} \) Now we can apply the quotient rule: \[ f'(x) = \frac{(e^x + e^{-x})(e^x + e^{-x}) - (e^x - e^{-x})(e^x - e^{-x})}{(e^x + e^{-x})^2} \] Now, let's simplify the numerator: \[ (e^x + e^{-x})^2 - (e^x - e^{-x})^2 \] Using the identity \( (a+b)^2 - (a-b)^2 = 4ab \), we can set \( a = e^x \) and \( b = e^{-x} \): \[ = 4(e^x)(e^{-x}) = 4 \] Thus, we have: \[ f'(x) = \frac{4}{(e^x + e^{-x})^2} \] Now, looking at the options provided: \[ \begin{array}{llll} \text{A. } \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}} & \text{B. } \frac{t}{(e^{x}+e^{-x})^{2}} & \text{C. } \frac{e^{x}}{(d^{x}+e^{-x})^{2}} & \text{D. } \frac{2}{(e^{x}+e^{-x})^{2}} \\ \end{array} \] None of the options match \( \frac{4}{(e^x + e^{-x})^2} \) directly. However, we can express \( \frac{4}{(e^x + e^{-x})^2} \) as \( \frac{2 \cdot 2}{(e^x + e^{-x})^2} \), which is equivalent to \( 2 \cdot \frac{2}{(e^x + e^{-x})^2} \). Thus, the closest option is: \[ \text{D. } \frac{2}{(e^{x}+e^{-x})^{2}} \] So the answer is: \[ \text{D. } \frac{2}{(e^{x}+e^{-x})^{2}} \]