## University Calculus: Early Transcendentals (3rd Edition)

${dy}={\frac{-1}{|e^{(-x)}|{\sqrt{((e^{-x})^2-1})}}}\times{{(e^{(-x)})}}{dx}$
We are given: $y={\sec^{-1}{(e^{-x}})}$ on differentiating both sides: ${\frac{dy}{dx}}={\frac{d({\sec^{-1}{e^{-x}}})}{dx}}$ on applying the chain rule: ${\frac{dy}{dx}}={\frac{1}{|e^{(-x)}|{\sqrt{((e^{-x})^2-1})}}}\times{\frac{d(e^{(-x)})}{dx}}$ ${\frac{dy}{dx}}={\frac{1}{|e^{(-x)}|{\sqrt{((e^{-x})^2-1})}}}\times{{-(e^{(-x)})}}$ ${dy}={\frac{-1}{|e^{(-x)}|{\sqrt{((e^{-x})^2-1})}}}\times{{(e^{(-x)})}}{dx}$ The final answer is: ${dy}={\frac{-1}{|e^{(-x)}|{\sqrt{((e^{-x})^2-1})}}}\times{{(e^{(-x)})}}{dx}$