## Calculus: Early Transcendentals 8th Edition

LHS $=\frac{d}{dx}_4\sqrt{\frac{1+\tanh x}{1-\tanh x}}$ $=\frac{1}{2}{e^{\frac{x}{2}}}$ =RHS (as needed)
LHS $=\frac{d}{dx}_4\sqrt{\frac{1+\tanh x}{1-\tanh x}}$ $=\frac{d}{dx}(\frac{1+\tanh x}{1-\tanh x})^{\frac{1}{4}}$ Using the chain rule: LHS $=\frac{d(\frac{1+\tanh x}{1-\tanh x})^{\frac{1}{4}}}{d\frac{1+\tanh x}{1-\tanh x}} \times\frac{d\frac{1+\tanh x}{1-\tanh x}}{dx}$ $=\frac{1}{4(\frac{1+\tanh x}{1-\tanh x})^{\frac{3}{4}}} \times\frac{d\frac{1+\tanh x}{1-\tanh x}}{dx}$ $\DeclareMathOperator{\sech}{sech}$ Using the quotient rule: LHS $=\frac{1}{4(\frac{1+\tanh x}{1-\tanh x})^{\frac{3}{4}}} \times\frac{\sech^2x(1-\tanh x)-(-\sech^2x)(1+\tanh x)}{(1-\tanh x)^2}$ $=\frac{1}{4(\frac{1+\tanh x}{1-\tanh x})^{\frac{3}{4}}} \times\frac{\sech^2x-\sech^2x\tanh x+\sech^2x+\sech^2x\tanh x}{(1-\tanh x)^2}$ $=\frac{1}{4(\frac{1+\tanh x}{1-\tanh x})^{\frac{3}{4}}} \times\frac{2\sech^2x}{(1-\tanh x)^2}$ Recall that: $$\sech^2 x=1-\tanh^2 x$$ Thus, LHS $=\frac{1}{2(\frac{1+\tanh x}{1-\tanh x})^{\frac{3}{4}}} \times\frac{(1-\tanh x)(1+\tanh x)}{(1-\tanh x)^2}$ $=\frac{1}{2}(\frac{1-\tanh x}{1+\tanh x})^{\frac{3}{4}} \times\frac{1+\tanh x}{1-\tanh x}$ $=\frac{1}{2}(\frac{1-\tanh x}{1+\tanh x})^{\frac{3}{4}} \times\frac{1+\tanh x}{1-\tanh x}$ $=\frac{1}{2}(\frac{1+\tanh x}{1-\tanh x})^{\frac{1}{4}}$ Recall that: $$\tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}$$ Hence, LHS $=\frac{1}{2}(\frac{1+\frac{e^x-e^{-x}}{e^x+e^{-x}}}{1-\frac{e^x-e^{-x}}{e^x+e^{-x}}})^{\frac{1}{4}}$ $=\frac{1}{2}(\frac{e^x+e^{-x}+e^x-e^{-x}}{e^x+e^{-x}-e^x+e^{-x}})^{\frac{1}{4}}$ $=\frac{1}{2}(\frac{2e^x}{2e^{-x}})^{\frac{1}{4}}$ $=\frac{1}{2}(e^{2x})^{\frac{1}{4}}$ $=\frac{1}{2}{e^{\frac{x}{2}}}$ =RHS (as needed)