The question is incomplete but we can find derivative here.
"y=ln(e^x+e^{-x})\n\\\\\\Rightarrow \\dfrac{dy}{dx}=\\dfrac{d}{dx}(ln(e^x+e^{-x}))\n\\\\\\Rightarrow \\dfrac{dy}{dx}=\\dfrac{1}{e^x+e^{-x}}\\times \\dfrac{d}{dx}(e^x+e^{-x})\n\\\\\\Rightarrow \\dfrac{dy}{dx}=\\dfrac{1}{e^x+e^{-x}}\\times(e^x-e^{-x})\n\\\\\\Rightarrow \\dfrac{dy}{dx}=\\dfrac{e^x-e^{-x}}{e^x+e^{-x}}"
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