The first derivative of functions: #y^3=e^x ln(x^2-1)# is?

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Answer 1 · 2018-06-13T07:36:15

#dy/dx=(e^xln(x^2-1)^2+2xe^x)/(3(x^2-1)(e^xln(x^2-1))^(2/3))#

#d/dx[y^3]=d/dx[e^xln(x^2-1)]#

#d/dx[y^3]=d/dx[e^x]ln(x^2-1)+e^xd/dx[ln(x^2-1)]#

#d/dx=dy/dx*d/dy#

#dy/dxd/dy[y^3]=d/dx[e^x]ln(x^2-1)+e^xd/dx[ln(x^2-1)]#

#dy/dx3y^2=e^xln(x^2-1)+(2xe^x)/(x^2-1)#

#y=(e^xln(x^2-1))^(1/3)#

#dy/dx=(e^xln(x^2-1)+(2xe^x)/(x^2-1))/(3(e^xln(x^2-1))^(2/3))#

#dy/dx=(e^xln(x^2-1)^2+2xe^x)/(3(x^2-1)(e^xln(x^2-1))^(2/3))#