Practice problem
Find the derivative. y=2csc−1xsec−1xy=2^{\csc^{-1}x}\sec^{-1}xy=2csc−1xsec−1x
y′=2csc−1x(ln(csc−1x)sec−1x+1xx2−1)y'=2^{\csc^{-1}x}\left(\ln(\csc^{-1}x)\sec^{-1}x+\frac{1}{x\sqrt{x^2-1}}\right)y′=2csc−1x(ln(csc−1x)sec−1x+xx2−11)
y′=2csc−1x[1−ln(csc−1x)sec−1x]xx2−1y'=\frac{2^{\csc^{-1}x}[1-\ln(\csc^{-1}x)\sec^{-1}x]}{x\sqrt{x^2-1}}y′=xx2−12csc−1x[1−ln(csc−1x)sec−1x]
y′=2csc−1x[ln(1xx2−1)+1xx2−1]y'=2^{\csc^{-1}x}\left[\ln\left(\frac{1}{x\sqrt{x^2-1}}\right)+\frac{1}{x\sqrt{x^2-1}}\right]y′=2csc−1x[ln(xx2−11)+xx2−11]
y′=1xsecxx2−1−ln2xx2−1y'=\frac{1}{x\sec x\sqrt{x^2-1}}-\frac{\ln 2}{x\sqrt{x^2-1}}y′=xsecxx2−11−xx2−1ln2
Type a concept, like “slope” or “photosynthesis”.
