If f(x) = frac{g(x) + g(-x)}{2} + frac{2}{[h( | vedclass

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(D) Given the function: $f(x) = \frac{g(x) + g(-x)}{2} + \frac{2}{[h(x) + h(-x)]^{-1}}$.Simplifying the expression,we get: $f(x) = \frac{g(x) + g(-x)}

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(D) Given the function: $f(x) = \frac{g(x) + g(-x)}{2} + \frac{2}{[h(x) + h(-x)]^{-1}}$.Simplifying the expression,we get: $f(x) = \frac{g(x) + g(-x)}{2} + 2[h(x) + h(-x)]$.Differentiating both sides with respect to $x$ using the chain rule:$f^{\prime}(x) = \frac{g^{\prime}(x) - g^{\prime}(-x)}{2} + 2[h^{\prime}(x) - h^{\prime}(-x)]$.Now,substitute $x = 0$ into the derivative:$f^{\prime}(0) = \frac{g^{\prime}(0) - g^{\prime}(0)}{2} + 2[h^{\prime}(0) - h^{\prime}(0)]$.$f^{\prime}(0) = 0 + 2[0] = 0$.

If $f(x) = \frac{g(x) + g(-x)}{2} + \frac{2}{[h(x) + h(-x)]^{-1}}$,where $g$ and $h$ are differentiable functions,then find $f^{\prime}(0)$.

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