differentiate y=x^x^x
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apply log on both the side..
then log y = x^x log x
diff both side wrt x
1/y Dy/dx = x^x * 1/x + log x * d(x^x)/dx
now if you follow similar procedure
for d(x^x)/dx you will get x^x*(1+logx)
now substitute in the above equation then
1/y Dy/dx = x^x-1 + logx * (x^x*(1+logx))
hence Dy/dx = y [x^x-1+logx(x^x(1+logx))]
then log y = x^x log x
diff both side wrt x
1/y Dy/dx = x^x * 1/x + log x * d(x^x)/dx
now if you follow similar procedure
for d(x^x)/dx you will get x^x*(1+logx)
now substitute in the above equation then
1/y Dy/dx = x^x-1 + logx * (x^x*(1+logx))
hence Dy/dx = y [x^x-1+logx(x^x(1+logx))]
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