How would one differentiate the function x^(cosh(x))?
T Trenters4325 Junior Member Joined Apr 8, 2006 Messages 122 May 29, 2006 #1 How would one differentiate the function x^(cosh(x))?
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 May 29, 2006 #2 logarithmic differentiation, that's how. \(\displaystyle \L\\y=x^{cosh(x)}\) log of both sides: \(\displaystyle \L\\ln(y)=ln(x^{cosh(x)})\) Property of logs: \(\displaystyle \L\\ln(y)=cosh(x)ln(x)\) Differentiate(product rule) both sides: \(\displaystyle \L\\\frac{y'}{y}=sinh(x)ln(x)+\frac{cosh(x)}{x}\) Since \(\displaystyle \L\\cosh(x)=\frac{e^{x}+e^{-x}}{2}\) \(\displaystyle \L\\y'=(sinh(x)ln(x)+\frac{cosh(x)}{x})x^{\frac{e^{x}+e^{-x}}{2}\)
logarithmic differentiation, that's how. \(\displaystyle \L\\y=x^{cosh(x)}\) log of both sides: \(\displaystyle \L\\ln(y)=ln(x^{cosh(x)})\) Property of logs: \(\displaystyle \L\\ln(y)=cosh(x)ln(x)\) Differentiate(product rule) both sides: \(\displaystyle \L\\\frac{y'}{y}=sinh(x)ln(x)+\frac{cosh(x)}{x}\) Since \(\displaystyle \L\\cosh(x)=\frac{e^{x}+e^{-x}}{2}\) \(\displaystyle \L\\y'=(sinh(x)ln(x)+\frac{cosh(x)}{x})x^{\frac{e^{x}+e^{-x}}{2}\)
