if \( g(1)=4 f(2)=2, f^{\prime}(2)=g^{\prime}(1)=-1 \) \( h(x)=\frac{1}{f(x)}+g\left(\frac{2}{x}\right)+x \Rightarrow h^{\prime}(2)= \) ?

To find \( h'(2) \) for the function \[ h(x) = \frac{1}{f(x)} + g\left(\frac{2}{x}\right) + x, \] we'll differentiate \( h(x) \) with respect to \( x \) and then evaluate at \( x = 2 \). 1. **Differentiate each term:** - The derivative of \( \frac{1}{f(x)} \) is \( -\frac{f'(x)}{[f(x)]^2} \). - The derivative of \( g\left(\frac{2}{x}\right) \) using the chain rule is \( g'\left(\frac{2}{x}\right) \cdot \left(-\frac{2}{x^2}\right) \). - The derivative of \( x \) is \( 1 \). 2. **Combine the derivatives:** \[ h'(x) = -\frac{f'(x)}{[f(x)]^2} + g'\left(\frac{2}{x}\right) \cdot \left(-\frac{2}{x^2}\right) + 1. \] 3. **Evaluate at \( x = 2 \):** - Given: \( f(2) = 2 \), \( f'(2) = -1 \), \( g'\left(1\right) = -1 \) (since \( \frac{2}{2} = 1 \)). \[ h'(2) = -\frac{f'(2)}{[f(2)]^2} + g'\left(1\right) \cdot \left(-\frac{2}{2^2}\right) + 1. \] Substitute the known values: \[ h'(2) = -\frac{-1}{2^2} + (-1) \cdot \left(-\frac{2}{4}\right) + 1 = \frac{1}{4} + \frac{1}{2} + 1 = \frac{7}{4}. \] **Final Answer:** \(\boxed{\dfrac{7}{4}}\)