Question:
differentiate sinx^sinx^sinx^sinx.............................
Answer:
Let y = (((sinx)sinx )sinx) .......
=> y = sinxy
Take log on both side
log y = y * log sinx
=> (1/y)*dy/dx = (dy/dx)*logsinx + y*cosx/sinx
=> (1/y)*dy/dx - (dy/dx)*logsinx = y*cosx/sinx
=> (dy/dx)*(1/y - logsinx) = y*cotx
=> (dy/dx)*{(1 - ylogsinx)/y} = y*cotx
=> (dy/dx)*(1 - ylogsinx) = y2 *cotx
=> dy/dx = (y2 *cotx)/(1 - ylogsinx)
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