the function of f is f(x)= ln(e^x + e^-x) using composite and quotient rules give an expression for f ' (x) and f ''(x) im confused about these rules....isnt f ' (x) just 1/e^x + e ^-x)????
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Let $\displaystyle f(u)=\ln u$ and $\displaystyle u=e^x+e^{-x}$ $\displaystyle \frac{df}{dx}=\frac{df}{du}\frac{du}{dx}=\frac{1}{ u}(e^x-e^{-x})=\frac{1}{e^x+e^{-x}}(e^x-e^{-x})$
does that mean that f ' (x) using the rules is therefore 1 ______ (e^x - e ^-x)^2
No. The answer is given above.
The most compact form is $\displaystyle \tanh(x)$
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