derivative of 24*sec^-1(x/16)
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Apr 2, 2007 #2 Re: integrals Hello, chadsgirl! You need to know: \(\displaystyle \L\:\frac{d}{dx}(\sec^{-1}u) \:=\:\frac{du}{u\sqrt{u^2\,-\,1}}\) Differentiate: \(\displaystyle \:24\cdot\sec^{-1}\left(\frac{x}{16}\right)\) Click to expand... We have: \(\displaystyle \L\:y \:=\:24\cdot\sec^{-1}\left(\frac{x}{16}\right)\) Then: \(\displaystyle \L\:y' \;=\;24\,\cdot\,\frac{\frac{1}{16}}{\frac{x}{16}\sqrt{\left(\frac{x}{16}\right)^2\,-\,1}} \;=\;\frac{24}{x\sqrt{\frac{x^2\,-\,256}{256}}}\;=\;\frac{24}{x\frac{\sqrt{x^2\,-\,256}}{16}}\) Therefore: \(\displaystyle \L\:y' \;=\;\frac{384}{x\sqrt{x^2\,-\,256}}\)
Re: integrals Hello, chadsgirl! You need to know: \(\displaystyle \L\:\frac{d}{dx}(\sec^{-1}u) \:=\:\frac{du}{u\sqrt{u^2\,-\,1}}\) Differentiate: \(\displaystyle \:24\cdot\sec^{-1}\left(\frac{x}{16}\right)\) Click to expand... We have: \(\displaystyle \L\:y \:=\:24\cdot\sec^{-1}\left(\frac{x}{16}\right)\) Then: \(\displaystyle \L\:y' \;=\;24\,\cdot\,\frac{\frac{1}{16}}{\frac{x}{16}\sqrt{\left(\frac{x}{16}\right)^2\,-\,1}} \;=\;\frac{24}{x\sqrt{\frac{x^2\,-\,256}{256}}}\;=\;\frac{24}{x\frac{\sqrt{x^2\,-\,256}}{16}}\) Therefore: \(\displaystyle \L\:y' \;=\;\frac{384}{x\sqrt{x^2\,-\,256}}\)
